Motion in a Straight Line (Revised)

7/30/2019 Motion in a Straight Line (Revised)

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M OTION IN A STRAIGHT LINE M ot ion is change in posi t ion o f an ob ject w i th t ime. The descr ip t ion o f how an ob ject m oves is

cal led kinematics. The study of forces and their effect on motion is cal led dynamics. The stra ight l ine

m ot ion or m ot ion in one d imension is the s imp lest m ot ion . Th is is also know n as rect i l inear m ot ion . W e

consider the motion of point objects cal led part ic les. I f the distance moved by an object is much larger

than t he i ts size, i t can be t reated as point object.The rect i l inear m otio n m ay be hor izontal , vert ical or slant.

Test your know ledge:1. In w hich of the fo l low ing examples, the object can b e considered a point o bject?

a . A cr icket bal l throw n by a f ie lder f rom the boun dary to t he wicket -keeper .b . A mo tor-bike moving betw een Ghaziabad and Chandigarh.c. A footbal l k icked by a player tow ards another player 4 m aw ay.d . The dr iver of a car dr iv ing i t in a larg e c ircular path.e. A cup fal l ing of f edg e of a tab le.

Position, Path Length a nd Displacem ent:

As mot ion is change in pos i t ion o f an ob ject w i th t ime, we need to spec i fy pos i t ion . For th is

purpose, a reference point and a set of coordinate axes are required. General ly we choose or ig in of arectangular coordinat e system as the r eference point . This coord inate system has three axes- X-, Y-, and

Z- axes. These are m utu al ly perpendicular axes. The posi t ion o f an object w i th r espect t o t h is coor dinate

system is descr ibed by t he coordin ates (x, y. z).

A part ic le is said to be at rest i f a l l the three coordinates x, y and z remain unchanged as t ime

passes. I f any one or more coordinates change with t ime,

the ob ject (part ic le) is said to be in m otio n.

We can choose a set of axes in a frame of

reference depending upon the si tuat ion given. For

example , to descr ibe mot ion in one d imension , on ly one

axis is required. It may be X-, Y- or Z- axis. For the

descr ip t ion o f m ot ion in tw o o r th ree d im ensions, a se t o f tw o or th ree axes, respect ively is requ i red .

The descr ip t ion o f m ot ion depends on the f rame o f re f e rence chosen. For exam ple , suppose weare t rave ll ing in a bus moving on a road. W i th respect t o a f ram e o f re fe rence a t tached t o t he ground,

bus is m oving. But w i th respect to a fram e of reference attached to us, the bus is at rest.

To describe motion along a straight line, an axis (X-, Y- or Z-axis) is so chosen that it coincides

w ith the path of th e par t icle . The pos i t ions o f the ob ject a re m easured w i th re fe rence to an or ig in . The

posi t ions of t he part ic le m ay be posi t ive or negative. For exam ple, on th e X-axis, posi t ion s to the r ight of

or ig in are assumed t o be posi t ive and t o th e left o f or i g in, as negative.

Path len gth (Distan ce):

I t is equal to t he actual path t raversed by a part ic le betw een the in i t ia l and f inal posi t ions. I t is a

scalar quanti t y. I t is a lw ays posi t ive. As shown in f igure 1, a part ic le star t s from or ig in O (x = 0) at t = 0

and A, B and C are i ts posi t ions at d i f f erent instant s of t im e. I f the part ic le m oves from O to B, thedistance covered by the

particle is OB = +140 m. This

distance is the path length t raversed by the part ic le.

Now, i f the par t ic le moves

back from B to C. The tota l

pa th length t raversed by the

part ic le is OB + BC = + 140 m + (+180 m ) = + 320 m . Tw o point s should be no ted h ere:

W hat is a f ram e of re ference?A coordinate system along w ith a

t im e-me asuring device (a clock) is called a

f ram e of reference.


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1. Path length (d istance) does no t depend on the d i rect ion o f m ot ion o f the par t ic le , i.e. it is a scalarquant i ty , a quant i t y tha t has a magn i tude on ly and no d i rect ion .

2. Path length is a lw ays posi t ive.Displacement:

The displacement of a part ic le is d i f fer ence betw een t w o posi t ions i t occupies. A movin g part ic le

continu ously changes i ts posi t ion. A change from one po sit ion x1 ( in it ia l posi t ion a t po in t A) t o anoth erposi t ion x2 ( f inal posi t ion at point B) is called a displacement . See figu re 2. It i s given b y

x= x2x1Displacement may be posi t ive, negative or zero. If x2 > x1, x is posi t ive; and i f x2 < x1, x is

negative. I t is zero i f x2 = x1.

Displacement is a vector quanti ty as i t has both

m agn i tude and d i rect ion . x is a vector d irected f rom x1 t o x2. In

a st ra igh t l ine mo t ion (o r rect i l inear m ot ion or o ne-d im ensiona l

mot ion) , there are on ly two possib le d irect ions. I f the given

direct ion of motion of a part ic le is assumed to be posi t ive, thed i rect ion opposi te to i t w i l l be negat ive . In the f igure 1 ,

d isp lacement o f t he par t ic le in m oving f rom O to B is g iven byx= x2x1

= (+140 m ) 0 m

= +140 m

The displacement has a magnitude of 140 m and is d irected along the + X-axis. If we calculate

the displacem ent of the part ic le from B to A, i t is g iven by

x= x2x1= (+ 80 m) 140 m

= 60 m.

Th e ve sign refers to the direct ion of d isplacement vector. This is an important fact about thest ra igh t l ine mo t ion . W e dont need to use vector notat ion in one-dimensional m ot ion . The positive

and negative signs are suff ic ient t o indicate the dir ect ion of m otio n of a part ic le.I f w e sta r t f rom a par t icu la r po in t and t hen m ove east 5 m eters, our d isp lacem ent is 5 m east . I f

we then tu rn back, w i th a d isp lacement o f 5 m west , we wou ld have t rave l led a to ta l d is tance (pa th

length) of 10 m , but o ur net d isplacem ent is zero, because in i t ia l and f inal posi t ions are sam e.

Thus, it is clear that the magnitude of d isplacement

may or may not be equal to the path length t raversed by an

object and the m agn i tude o f the d isp lacem ent f o r a course o f

mot ion may be zero bu t the cor respond ing pa th length is no t

zero .

Test your know ledge:2. A man leaves his home in his car goes to store and then returns to his home after travelling 5 km in one hour. What is

his distan ce and displacem ent?3. An athlete runs on a circular path of radius r and com pletes half the revolution. What is the displa cem ent an d t he

distance t ravel led by the a thlete?4. U n d e r w h a t c o n d i t i o n t h e d i s t a n c e c o v e r e d b y a p a r t i c l e i s e q u a l t o t h e m a g n i t u d e o f

d isplacement of p ar t ic le?5. Can a bod y he sa id to be a t res t as we l l as in m ot i on a t t he sam e t ime?

Exam ple 1 A body moves towards east by a distance 3 km and turns towards north and moves a distance of 4 km.

W hat is the displacem ent of the body?

The num erical rat io of m agnitude

of displacem ent t o dista nce is 1 . The

value o f this rat io is equa l to 1 in a

uni form m otion a long a stra ight l inein a given direction.


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Position-time gr aph:

The motion of a part ic le in one-dimensional motion can be descr ibed by a posi t ion-t ime graph.

We choose an axis ( for example, X-axis) a long the direct ion of motion of the part ic le. As the motion is

one-dim ensional , only the x-coordinate var ies wi th t im e. The resul t ing graph is x-t graph. The x-t graph

m ay take d if fe ren t shapes depend ing on the type o f m ot ion .

The f igure 3 shows posi t ion-t ime graphs of d i f ferentpart ic les A, B, C and D. The p art ic le A has th e sam e po sit ion (10 m

from the reference point) at a l l t imes. Hence, i t is stat ionary. The

part ic les B, C and D show di f feren t d isplacement s of the p art ic le at

d i f ferent t imes. Hence, these are a l l in motion. But they represent

d i f fe ren t types o f m ot ion .

In posi t ion-t ime graphs of part ic les B and C, the angle

between the stra ight l ine and the X-axis (i.e. , the slope of the

posi t ion-t ime l ine) is a lways the same at their respective values

(di f ferent b ut constant ) . In other w ords, they cover equal d istances

in equal intervals of t im e. Therefo re, they are said to be in uniform

mot ion along a stra ight l ine.Th e imp ortant d i f ference betw een t he m ot ions of par t icles B and C is that they are m oving in

oppo site direct ions. I t is clear fro m the posi t ion-t im e graphs of part ic les B and C. The posi t ion o f part ic le

B is increasing from the reference point (or or ig in) a long +ve X-axis but the posi t ion of part ic le C is

decreasing from t he reference point.

For part ic le D, the posi t ion-t ime graph indicates that in equal t ime- intervals, the posi t ion

changes by di f f erent amou nt s. Hence, i t has var iable m ot ion . The direct ion of m otio n of part ic les B and

D is sam e (alon g +X-axis).NOTE:

There is no accelerat ion in the motion of part icles B and C . So, no force is required t o make the

part ic les move with uni form veloci ty ( to be discussed later) . In such cases, instantaneous veloci ty is

always equal to th e average veloci ty. The m otion of part icle D is accelerate d .

Test your know ledge:6. W hat is the nature of t he displacement- t ime curve of a body m oving wi th constant veloci ty?7. W hat is the slope of the displacement-t im e curve, when the body m oves w ith constant velocity?8. The displacement of a m oving body is propor t ional to th e square of t ime. Is the body m oving wi t h uni form

veloci ty or uni form accelerat ion?

9. W hat is the nature of d isplacement - t ime curve of a body m oving w i th uni form accelerat ion?Exam ple 2

A drunkard w alking in a na rrow lane takes 5 steps forw ard and 3 steps backw ard, fo l low ed aga in by 5

steps forw ard a nd 3 steps backwa rd, and so on. Each step is 1 m long an d requires 1 s. Plot the x-t gra ph of his

m ot ion. Determine graphical ly and otherw ise how long the drunkard takes to fa l l in a p i t 13 m aw ay from t he

start .

Aver age Velocity and Aver age Speed:

Average veloci ty m easures the rat e at w hich the posi t ion o f a part ic le changes w ith t im e and in

w hat d irect ion . I t is def ined as the rat io of change in po sit ion (or d isplacement x) and the t im e in te rva ls

(t) , in which the displacement occurs. Again, we dont need to use the vector notat ion in one-

dimen sional mot ion . It can be specified by + ve and ve signs. Thus, th e average velocit y is given by

= =


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w h e r e x2 an d x1 are the posi t ions of the part ic le at

ins tan ts o f t ime t2an d t1 , respectively. The bar over a symbol is a standard notat ion that refers to an

average quanti ty. Hence, indicates the average veloci ty.Th e SI unit for velocity (both average and instantaneo us) is m / s o r m s1 . The SI unit f or speed

(both average and instantaneo us) is also m / s o r m s1 . Their dim ensions are [LT 1] .The average velocity i s a vector quant i ty l ike d isplacem ent.

Th e average velocity can be po sit ive or negat ive depen ding upon th e sign of the displaceme nt .

It can also b e zero i f t he displacem ent is zero. In f igure 3, th e x- t graphs of th e part ic le A show s that i t is

stat ionary. The x- t graphs o f th e par t ic le B show s tha t i t i s moving wi t h constan t p os it i ve ve loci ty and t he

part ic le C is moving with constant negative veloci ty. The x- t graphs of the part ic le D shows that i t is

m oving wit h var iable posi t ive veloci ty.

As w e know t he m agn i tude o f d isp lacem ent m ay be d i f fe ren t f rom the actua l path length . The

t ime rate of change of the actual path is called average speed . I t is def ined as the rat io of tota l path

length t ravel led and t he to ta l t im e interval in w hich th is path length is covered. Thus,

Average speed=

We cannot ascer ta in the d i rect ion o f mot ion f rom the average speed o f a par t ic le . We are

concerned on ly w i th t he pa th length . Thus, i t i s alw ays posit ive . I n one-dimen sional mot ion of a part ic lealong a given direct ion, the m agnitude of d isplacem ent is equal to t he to ta l path lengt h. In th is case, the

magnitude of average veloci ty is equal to the average speed. This wi l l not be true i f the direct ion

changes.

For example , refe r to f igure 1 . If the par t icle mo ves f rom O to A and com es back to O in the

sam e t im e interval of 10 second each, then the average speed is 8 m/ s but t he average veloci ty is zero

(as displacem ent becom es zero).

NOTE:1. I f a part ic le travels a long a stra ight l ine wit h speed v1 for th e distance x1 and then w i th speed v2

for d istance x2, then i t s average speed dur ing the w hole jour ney is given by

vav () = Or vav = = =

(

)

I f the t w o distances are equal , i.e., s1 = s2, th is re lat ion gets reduced to

vav =

Here, vav is equal to t he ha rmonic m e a n o f v1 an d v2.

The above discussion can be extended to a journey having m ore t han tw o part s as:

vav =

2. I f a part ic le travels a long a stra ight l ine wit h speed v1 f o r t i m e t1 and then w i th speed v2 f o r t i m et2, then i t s average speed dur ing the w hole journ ey is g iven by

vav =

Or vav = = I f the tw o t im e dura t ions are equal , i.e., t1 = t2, th is re lat ion gets reduced to

vav =

Here, vav is equal to t he ar i thmet ic m ean of v1 an d v2 .

The above discussion can be extended to a journey having m ore t han tw o part s as:

vav =


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3. A part ic le having zero speed cannot have non-zero velocity because veloci ty has both them agnitude (speed) and direct ion .

Test your know ledge:10. Is i t po ssible tha t t he average veloci ty of a p art icle is zero w hen i ts average speed is not zero?

Exam ple 3A car has to cover a d istance of 60 km . I f hal f of the tota l t im e i t t ravels w ith a speed of 80 km/ h and in

rest hal f t im e, i ts speed becom es 40 km/ h. W hat is the a verage speed of the car?

Instantaneous Velocity and Instantaneous Speed:

The average ve loc i ty o f a par t ic le does no t show how fast i t m oves a t d i f fe ren t instan ts o f t ime

dur ing a g iven in te rva l o f t ime. A phys ica l quant i ty tha t p rov ides th is in fo rmat ion is instantaneous

velocity (or simply veloci ty) vat an instant t.

The instantaneous veloci ty at a t im e t is def ined as th e l imi t o f the average veloci ty as the t im einterval t becom es inf in i t esim al ly sm al l . Thus,

v= =

w he re the symbo l stands for t akinglimit as t 0 ( tends to zero) o f the quant i ty on i tsr igh t ,

. The quant i ty

is the d i f fe ren t ia t ion o f x

w i th respect to t. I t i s the ra te o f change o f posi t ion w i th respect to t im e, at t he g iven instan t .

The veloci ty of the p art ic le at an instant can be calculated

bo th graphically an d numerically . As show n in f igure 4 (graphical

method), by the defin i t ion of the average veloci ty, the slope of

l ine AB gives the value of average veloci ty over the interval t1 t o

t2. I f the t ime interval is decreased from t3 t o t4, then l ine AB

becom es CD and i t s slope gives the value of the average veloci ty

over the in t e rva l t3t o t4. As the t im e interval appro aches zero (t

0) , the l ine AB becom es tangent to the pos it ion- t im e graph at

the po in t P and the veloci ty at instant t is g iven by the slope of

the tangent a t th e po in t and t h is is the instantaneous veloci ty of

the par t ic le a t the instan t t. This makes the meaning of the

l imit ing process qui t e clear.

The graphical method for calculat ing the instantaneous veloci ty of a part ic le is a lways not an

easy method. The instantaneous veloci ty at d i f ferent instants can be easi ly calculated i f an exact

expression for the posi t ion of the part ic le as a funct ion of t ime is g iven. For th is purpose, we can

di f feren tiate t he given expression t o calculate instantaneo us veloci ty at d i f ferent instants as fo l lows:

v=

Instan tane ous spee d is the m agnitude of veloci ty. I t should b e noted that tho ugh average speed

over a f in i te interval of t im e is greater or equal to the m agnitude of th e average veloci ty, instant aneous

speed at an instant is equal to t he m agnitude of th e instant aneous veloci ty at that instant .

Test your know ledge:11. W hat do es the speedomet er of a car measure?12. Can a bod y have constant speed and st i l l have varying velocity?13. Can a pa rt icle have con stant veloci ty and va rying speed?

Exam ple 4


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The position of an object m oving along x-axis is given by x = a + bt2

w here a = 6 .0 m, b = 5 .5 m s2

and t

is m easured in seconds. W hat is its init ial velocity a nd velocity at t = 4.0 s. W hat is the a vera ge velocity bet w een

t = 4 .0 s and t = 6 .0 s?

Exam ple 5A car moves along a straight l ine whose motion is given by x =

12 t2

4 t + 6 , w here x is in m etres and t is in seconds. W hat is its init ialvelocity?

Exam ple 6Discuss the motion of a particle having position-time graph as

shown in f igure 5.

NOTE:

1. Posit ion-t im e curve (or d isplacem ent - t im e curve) cannot be p aral le l to po sit ion axis as i t m eansa f in i te d isp lacement in no t ime. In o ther words, i t

denot es an inf in i te veloci ty.2. Displacement- t ime and distance-t ime graph cannot

be a closed curve in one dimensional motion (see

f igure 6) as i t means tha t the ob ject w i l l be a t twoposi t ions at the same instant of t ime. Also, the

reversal of t im e is not p ossib le. Arrow s on t he curves

are m eaningless.

3. The displacement- t ime graph cannot take sharp turns as i t g ives twodi f ferent veloci t ies at that point. Figure 7. Also, the curve wi l l not be

di f fer entiable at such point s.

4. The pos i t ion- t ime graph (d isp lacement - t ime graph) does no t show thetra jectory o f the par t ic le .

5. The distance-t ime graph is a lways an increasing curve for a movingobject as d istance never decreases with t ime. Therefore, the graph

shown in f igure 8 is not possib le.6. Th e instantaneous speed is always equal to the magnitude of

instantaneous velocity (not a lways true for their average values)

because for an arbi trar i ly smal l in terval of t ime, the magnitude of

d isp lacement is equa l to t he length o f the pa t h .

Velocity-time (v- t) gr aph:

The f igure 9 shows d i f fe ren t ve loc i ty - t ime graph A, B, C and D. We can draw the fo l lowing

conclusions from th ese curves:

1. Th e v-t graph is paral le l to t im e axis (graph A). The veloci ty isconstant and hence, accelerat ion = 0. This l ine on v-t graph is

equivalent to t he curve for part ic le B on x t graph o f f igure 3 .2. The graph is an obl ique stra ight l ine having posi t ive slope

(graph B). The m otio n in th is case has constant accelerat ion

as veloci ty increases at constant rate.

3. The graph is an obl ique stra ight l ine having negative slope(graph C). The m otio n in t h is case has constant retardat ion as

veloci ty decreases at constant rate but st i l l moving in

posi t ive d irect ion .


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4. Note that for graph B and C, the part ic les are moving in the same direct ion in one-dimensionalm otio n. One increasing (B), other decreasing (C).

5. I f a graph l ike C crosses the t ime axis, i ts d irect ion of m otio n changes ther eafter . This si tuat io n isexpla ined by graph E in f igure 10. Also note t hat t h is v-t graph (E) is corresponding t o x- t graph

of f igure 5 .

6. The graph is a curve of increasing slope (graph D). I t show sthat the p art ic le has var iable veloci ty w hich is increasing wit h

t im e. The m otio n is accelerated and t he accelerat ion goes on

increasing.

Exam ple 7Discuss the motion of a particle having velocity-time graph as

shown in f igure 11 (a lso refer to f igure 12 a fter discussion).

NOTE:

1. In uni for m m otio n, the average and instantaneous veloci t ies are a lways sam e at each point of i t spa th or a t each instan t .

2. In un i fo rm m ot ion , the ve loci ty is independent o f the cho ice o f t im einter val and or ig in.

3. I f the average veloci ty of a part ic le is zero in a t im e interval , then i t ispossib le that the instant aneous veloci ty is never zero in that interval .

I f a part ic le comes back to in i t ia l posi t ion, the average veloci ty is

zero but th e instant aneous veloci ty not zero.

4. An object can have constant speed but var iable veloci ty (as inun i fo rm c i rcu la r mot ion) bu t the opposi te is no t t rue , i.e., constant

veloci ty w i th var iable speed is not possib le.

5. A re a be t we e n v-t graph and t ime-axis can give both thedisplacement and dis tance in that t ime interva l . For

displacement, appropr iate sign has to be taken into account. For

distance is g iven by the t ota l area betw een v-t graph and t im e-axiswithout considering sign. See figure 13. Displacement is zero as +

ve area cancels out ve area. But d istance is equal to the sum o f

areas of t w o tr iangles.

6. A v-t graph cannot be a closed curve in one-dimensional motion.Th e v-t graph shown in f igure 14 is not possib le in one-dimensions

as the part ic le wi l l have veloci t ies in posi t ive as wel l as negative

direct ions. Also, arrow on t he curves is meaningless.

7. The veloci ty- t ime graph cannot take sharp turns. I t means theyveloci ty cannot change abrup tly.

8. A speed-t ime graph cannot be l ike a graph as shown in f igure 13because th e speed of a part ic le can never b e negative.

9. With he lp o f a g iven v-t graph, corresponding s-t graph and a-tgraph can be plott ed.

Acceleration:

The accelerat ion is def ined as the rate of change of veloci ty wi th t ime. Average accelerat ion

measures the ra te a t wh ich the ve loc i ty o f a par t ic le changes wi th t ime and in what d i rect ion . I t i s

defined as the rat io of change in velocity vand the time intervals t. Again, w e dont need to use the


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vector no tat ion in one-dim ensional mo t ion. I t can b e speci f ied by + ve and ve signs. Thus, th e averageaccelerat ion is given by

= =

w h e r e v2 an d v1 are the instan taneous veloci t ies of the part ic le at instants of

t i m e t2an d t1, respectively. The bar over a symbol is a standard notat ion that refers to an average

quant i ty. Hence, indicates the average accelerat ion. I t is the average change of veloci ty per uni t t im e.Th e SI unit o f accelerat ion is m s2 . I ts dim ensions are [LT 2] .The slope of t he veloci ty-- t im e graph gives the average accelerat ion in th e given t im e interval .

The instant aneous accelerat ion at a t im e t is def ined as the l im i t of th e average accelerat ion asthe time interval t becom es inf in i tesimal ly smal l . Thus,

a = =

Th e slope of the tangent to the ve loc ity - t ime graph at any instant t gives the accelerat ion atthat instant . This is the instantaneous accelerat ion of the part ic le at the instant t. The uni form

accelerat ion is equal to t he average accelerat ion over t he per iod in w hich i t rem ains uni fo rm .

It is now clear t hat accelerat ion is produ ced wh enever t he veloci ty changes. A change in a vector

quant i ty l ike veloci ty occurs w hen i t m agnitude (speed) or d irect ion or bot h change.

Like velocity, accelerat ion can also b e p osit ive, ne gat ive or zero . The curve D in posi t ion-t imegraphs of f igure 3 has posi t ive accelerat ion. B and C have zero accelerat ion. Figure 5 show s mot ion w ith

negative accelerat ion (retardation). Note that the posi t ion-t ime graph curves upward for posi t ive

accelerat ion; dow nw ard for n egative accelerat ion and i t is a stra ight l ine f or zero accelerat ion.

Since the topic under study is uni formly accelerated motion, any graph of accelerat ion versus

t ime must be a stra ight l ine paral le l to the t ime axis. Any other kind of accelerat ion- t ime graph is

beyond the scope of th is chapter. In th is case, the average accelerat ion equals the constant value of

accelerat ion dur ing the given inter val of t im e.

Th e veloci ty- t ime graphs for motions having di f ferent types of accelerat ion are a lready

discussed in t he f igure 9 and f igure 10.

Ar ea under v- t and a- t curves:

The area be tw een the ve loci ty - t ime graph and the t im e-ax is g ivesthe d isplacement in th e given interval of t im e. See f igure 15. It is clear from

the re la t ion

v=

From th is re lat ion

dx= v dt

The area between the accelerat ion-t ime graph and the t ime-axis

gives the velo ci ty in th e given interval of t im e. See f igure 16. I t is clear from

the re la t ion

a=

From th is re lat iondv= a dt

Test your know ledge:14. W hat is the nature of accelerat ion - t ime graph wh en the body moves wi th

constant accelerat ion?

15. W hat is the area und er the veloci ty - t im e curve in the case of a bodyprojected ver t ical ly upw ards f rom th e ground af ter reaching the ground?

16. W hat does the area of a cceleration-time graph represent ?


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17. As th e body m oves in a straigh t l ine, i t m oves sma l ler and sm al ler distances in equal interv als of t im e.W hat do i n fer about the mot i on?

18. I f a body m oves 8 m, 12 m , 16 m in successive second s, how do you d escribe i ts mo tion ?19. Can a b ody ha ve zero veloci ty and no n-zero accelerat io n? Explain w ith exam ples.20. A body ha s constant a ccelerat ion . Can i t t ravel oppo site to the accelerat ion, durin g i ts mot ion? Give an

example.21. I f a body m oves w i th uni form veloci ty , what is i ts accelerat ion? W hat w i l l be the nature of velocity- t im e

graph?

22. W hen w i l l a body have accelerated mot ion?23. Can a pa rt icle with zero accelerat ion speed up?

NOTE:

1. A part ic le can have accelerat ion without having any velocity . For example, a part ic le thrownvert ical ly up has zero veloci ty at t he highest poin t bu t a dow nw ard accelerat ion .

2. In one-dimension, a part ic le wi th constant speed must have zero accelerat ion. I t is not true inun i fo rm c ircu lar m ot ion- a tw o-d imensiona l mot ion . We can a lso say th a t it i s possible in circular

mo t ion tha t

0 (i.e. magnitude of accelerat ion 0) bu t

|

| = 0 ( i.e. rate of change of

speed = 0).

Exam ple 8A tra in passes a pole in 5 seconds and a br idge of the sam e length as the t r a in in 10 second. De scr ibe

t h e v e l o c it y a n d a cce l e r a t i o n o f t h e t r a i n ?

Kinematic equations for uniform ly accelera ted motion:

For uni formly accelerated motion, the average accelerat ion equals the constant (uni form)

accelerat ion of the m otion . Thus, i f u be t he in i t ia l veloci ty (at t = 0) and v the f ina l ve loci ty a f te r t im e t

o f a par t ic le moving wi th constan t (un i fo rm) acce le ra t ion a, then f rom the equat ion o f average

accelerat ion, w e get

a =

=

Or v= u+ a t

Figure 17 d epicts th is equation graphical ly.

The area between the veloci ty- t ime graph and the t ime-axis g ives

the d isp lacement in the g iven in te rva l o f t im e. Therefore, the displacem ent

xof t he part ic le is g iven by:

Area between instants 0 and t = Area of tr iangle PQR + Area of

rectan gle OPRS

Or x= (vu) t + ut

Putt ing (vu) = at, we get

x= at

2 + u t

Or x= ut + a t2 Here, the in i t ia l posi t ion o f th e part ic le is zero (i.e. x0 = 0).

Also, from x= (vu) t + u t, we get

x= t

And f rom v= u + a t, we get

t =

Thus,

v= u+ a t

x= ut +

a t2

v2 = u2 + 2a x


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x= t = =

Or v2 = u2 + 2 a x

W e can also get th is equation by el im inating t f r om the equa t ionsv= u+ a tan dx= ut + a t

2 .

Thus, we get thr ee kinemat ic equation s of rect i l inear m otio n for constant accelerat ion.

See th e text -box.

I f the in i t ia l posi t ion o f t he part ic le is x0 , the second and th i rd equat ions ge t m od i f ied as fo l lows

(See t he t ext-box):

Calculus M ethod:

From the defin i t ion of accelerat ion,

a =

Or dv= adt

In tegrat ing wi th in the l im i ts o f ve loci ty and t im e, we get

= = a Or vu = atOr v= u+ a t

From the de f in i t ion o f ve loci ty ,v=

Or dx= vd t

In tegrat ing wi th in the l im i ts o f ve loci ty and t im e, we get

= = ( + )

Or xx0 = ut + a t

2

Or x= x0 + ut+ a t

2

Also,

a= =

= v

Or vdv= a dx

In tegrat ing wi t h in the l im i ts o f ve loc i ty and pos it ion , w e ge t

= = a(xx0)

Or v2 = u2 + 2 a (x x0 )

The calculus m etho d can also be used for m otio n w ith no n-uni for m accelerat ion also.

Displaceme nt in the nth second :

Let xn be the displacem ent in n seconds an d xn- 1 be the displacem ent in th e f i rst (n-1) of th ese n

seconds. The displacem ent in t second s is given b y

x= ut + a t

2

Therefore, in n seconds,

xn = un + an2

And t he d isp lacement in (n-1) second s is given b y

xn1 = u(n1) +a(n1)

2

There fore the d isp lacement dur ing the nth second is obtained by

subtract ing xn1 f r om xn. See f igure 18. Thus,

xnt h = xnxn1

v= u+ a t

x x0 = ut + a t

2

Or x= x0 + ut +

a t2

an d v2 = u2 + 2 a (x x0 )


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Or xnt h = u +a (2 n 1 )

Exam ple 9 Figure 19 shows the time-displacement graph for a particle in one

dimension. Dra w the corresponding t ime-veloci ty graph.

Example 10

A car moving a long a stra ight h ighway with speed of 126 km h

1

isbrought to a stop w ithin a d istance of 200 m . W hat is the retardat ion of thecar (a ssumed uniform), and how long does i t take for t he car t o stop?

Example 11The veloci ty of a part icle changes from +3 m / s to +8 m / s. In w hat d irect ion, posi t ive or nega t ive, is the

part icle mo ving? Is the a ccelerat ion positive or nega tive? Are the velocity and acceleration in the sam e direction

during the t ime inter val of the change in velocity? Explain.

Ver tical Motion under Gr avity:

The value of accelerat ion due to gravi ty is assumed to be constant. Therefore, the kinematic

equation s can be appl ied.

D o w n w a r d m o t i on :

The acceleration a w il l be replaced

by g and the equat ions wi l l take the fo rm asshow n in f i rs t box.

U p w a r d m o t i o n:

The acceleration a w il l be replaced by g and the equ at ions w i l l take the fo rm as show nin second box.

NOTE:

1. I f a body is d ropped f rom a he igh t h, then t he speed w hen i t reaches the ground isv= [Pu t t ing u = 0 , x= h in v2 = u2 + 2 gx]

2. I f a body is d ropped f rom a he igh t h, then the t im e to reach the ground ist =

[Pu t t ing u = 0, x= h in x= ut + gt2]

3. I f a body is th row n upw ard wi th ve loci ty u, then the maximum he igh t H isH =

[Pu t t ing v= 0 , x= H in v

2 = u2 2gx]

4. I f a body is th row n upw ard wi t h ve loci ty u, then the t im e in wh ich i treaches the m aximu m he igh t is

t = [Pu t t ing v= 0, and a = g in v= ugt]

5. I f a body is d ropped f rom som e he igh t , then t he ve loci ty a t ta ined byi t a f te r t ime t is given by (Using v = u + gt an d tak ing upward

direction as positive )

v= gt

A v-t graph is a stra ight l ine as show n in f igure 20.6. I f a body is d ropped f rom some he igh t , then the d is tance covered by i t in t ime t is given by(Using x= ut +

a t

2 an d taking upw ard direct ion as posit ive)

x= a t

2

A x-t graph is a parabol ic curve as show n in f igure 21.

v= u + gt

x= ut +

gt2

v2 = u2 + 2 gx

v= u gt

x= ut

gt2

v2 = u2 2 gx


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7. I f a body is th row n ver t ical ly up wi t h ve loc ity u f rom a he igh t h(o r d ropped a t a he igh t h f r om a ba ll oon mov ing upward w i th

veloci ty u) , then the t ime taken by the body to reach the

ground can be ca lculated in tw o w ays(Figure 22).

In the f i rst m ethod , the t o ta l t im e is calcu la ted in tw o

par ts- From po in t o f p ro ject ion B t o t he h ighest po in t C ( le t i t

be t1) and f rom po in t C to the ground ( let i t be t2). The velocit y

at C becom es zero.

The t im e t1 can be calculated u sing equation v= u + a t

and pu t t ing v= 0 , a =g an d t= t1. This give s t1=.

To calculat e t2 ( t ime t aken f rom po in t C to t he ground), we

fi rst calculate t he distance BC ( let i t be h ) using equation v2 = u2 +

2axand put t ing v= 0, a =g an d x= h . This gives h =. Now, t2 is

the t im e taken by the body to cover h + h . I t can b e calculated u sing

the equat ion x= u t + at

2 and put t ing u = 0 , a =g, t= t2 an d x= h+

h = h + . Form th is, t2 = .In the second method , we can calculate the tota l t ime in a

sing le step . W e consider the d isp lacem ent o f the bo dy be tw een the

in i t ia l posi t ion (point of p roject ion B) and t he f inal posi t ion (ground)

on ly . We do no t have to cons ider the pa th o f the mot ion as the

mot ion is under constan t acce le ra t ion I f we take the downward d i rect ion as pos i t ive , then

displacement x= + h an d a = g but in i t ia l ve loc i ty u shou ld be t aken as negative as i t is in upw ard

direct ion. Thus, from x= u t + at

2, we get

h = ut + gt

2

Solving th is equation, w e get th e tot a l t im e. This met hod is a lso very useful in solving sim i lar

prob lem s in Pro ject i le M ot ion .Example 12

A bal loon is r ising vert ical ly upwa rds with uni form speed of 10 m / s. W hen i t is 400 m above the ground, a

stone is dropped from i t . Af ter how much t ime and with what veloci ty wi l l the stone hi t the ground? (g = 10m / s

2)

8. I t shou ld be no ted tha t the speeds o f the body a t po in ts B and D areequa l . Also , the t im e taken to reach f ro m po in t B t o po in t C is equa l to

t im e taken to m ove f rom po in t C to D, i f w e neg lect the a i r resistance.

9. I f retarding force due to a ir resistance is considered, then a body t a k e slesser t ime to reach the highest point an d la rger t ime to reach the

ground as compared to that taken in the absence of air resistance . In

the f i rst case, effect ive value of g increases as air resistance acts

downward. In the second case, effect ive value of g decreases as airresistance acts upw ard.

10 .A body th row n upw ard w i th ve loci t y u is at the same height h a t t w oinstants of t ime ( let t1 an d t2) , once wh i le go ing up and then coming

dow n as show n in f igure 23 . Under th is cond i t ion , w e can prove tha t h =gt1t2 an d u =

g (t1 +

t2) . In th is case, the equ ation r e lat ing h an d u is h = ut + gt

2. This quadrat ic equation has tw o

root s, let t1 an d t2. Now f r om the proper t ies o f roo ts , we can w r i te tha t


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t1 + t2 = = (Com pare wi th ax2 + bx+ c= 0)This gives

u=g (t1 + t2)

an d

t1t2 = = This gives

h=gt1t2

I t can a lso be pro ved tha t the t ime t aken to reach the h ighest

po in t is t = (t1 + t2 )

11.I f th ree ba l ls a re th rown wi th same speed f rom a g iven he igh t invacuum (or i f a i r resistance neglected) such that one is thrown

ver t ica l ly downward , the second ver t ica l ly upward and the th i rd

hor izontal ly (Figure 24), then al l the three bal ls reach the ground

w i t h same speed . In th is case, the bal ls need not be

identical .Test your know ledge:

24. A man throws two bal ls f rom the top of a bui ld ing- one bal l s t raight up wi th an in i t ia l speed u and other bal l s t raight

down wi th the same in i t ia l speed. Which bal l has larger

speed w hen i t h i ts the ground?

25. A ball is dropped from the top of a building. Another ball isthrown horizontal ly at the same instant from the same point.

W hich bal l w i l l h i t the groun d f i rs t?

Example 13 A ball is thrown down vertically with an init ial speed of

20.5 m / s f rom a height of 58.8 m . (a) W hat w i ll be its speed just

before i t s tr ikes the ground? (b) How long w i ll i t ta ke for the ba l l to reach the ground? (c) Wha t w ould be the a nswe rs to (a ) and (b) i f

the bal l were throw n direct ly up f rom the sam e height and w ith thesame in i t ia l speed? (d) How long wi l l i t take for the bal l to reach

ma x imum he igh t? Galileos law of od d num bers (Distan ces covered in equ al interval of t im e) :

The distances covered by a part ic le, star t ing from rest (both in hor izonta l and ver t ica l m ot ions),

in equal inter vals of t im e are in the rat io 1: 3:5 - - - etc.

This can be proved as fo l low s:

First w e calculate t he distances covered in t im e intervals t, 2 t, 3 t, 4 t, - - -, using th e equatio n x= ut +

a t2 =

a t2 (u = 0 ) . Clearly, t hese distances are

xt= a t2 ,x2 t=

a (2 t)

2 = 4 ,x3 t=

a (3 t)

2 = 9 ,x4 t=

a (4 t)

2 = 16 , and so on .Clear ly, the distances covered in the f i rst in t erval t, f i rst t w o 2 t, f i rst t hree 3 t, - - - et c. are in the

rat io 1 : 4 : 9 : 16 : - - - etc.

Stopp ing distan ce of ve hiclesI f the stopping distance for a

vehicle of mass m moving with speed

u is x, then for a speed nu, the

stopping distance becomes n2

x ( for

the same retardat ion a). It is clear

from the following analysis.

The final spee d is zero in both th e

cases. U sing v2

= u2

+ 2a x in the firstcase, w e get

0 = u2 2a x( for retardat ion)

Or u2

= 2a x

In t he second case,0 = (nu)

2 2a x

Or n2u

2= 2a x

From t he tw o equat ions, we get

x = n2x


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Now , the distance covered in the f i rst in ter val t is given by

x1 = xt=a t

2

The distance covered in t he second inter val t is given by

x2 = x2 t xt= 4 a t2 = 3 The distance covered in the t h ird int erval t is given b y

x3 = x3 t x2t = 9 4 = 5 Clear ly, the d istances covered by a p art ic le, star t ing fro m rest, in equal intervals of t im e are in the

rat io 1 : 3 : 5 - - - etc .

Example 14

A car m oving at a speed of 10 m / s is stopped over a d istance of 30 m af ter brakes are appl ied. W hat w i ll

be the dista nce i f i t is m oving at 30 m / s?

Relative Velocity:

In comparison to re lat ive veloci ty in tw o-dim ensions and three-dim ensions, the re lat ive veloci ty

in one-dimension is simpler. Consider two bodies P an d Q moving wi th un i fo rm ve loc i t ies vP an d vQ

paral le l to x-ax is, bo th w i th re fe rence to the ground. Le t t he in i t ia l pos it ions (at t im e t = 0) o f bod ies Pan d Qbe xP(0) and xQ(0) , respectively. The in i t ia l d isplacem ent f rom body Pto body Q is given b y

xQP (0) = xQ (0 ) xP(0 )The posi t ions o f the tw o bod ies at t ime t are, respectively, given by

xP(t) = xP(0) + vPt

xQ(t) = xQ (0) + vQ t

The d isp lacement f rom body Pto body Qa t t ime t is given b y

xQP (t) = xQ(t) xP(t)

Or xQP (t) = [xQ(0 ) xP(0)] + (vQvP) t

I t i s c lear f rom th is equat ion tha t w i th respect t o the body P(or as seen from body P) , body Q has aveloci ty vQ vPbecause the displacement from P t o Q changes un i fo rm ly by an am ount vQ vP in each

un i t o f t im e. In o ther word s, the velocity of bod y Q re la t ive to body Pis vQ vP.

Thus, the veloci ty of body Q re la t ive to body Pis given b yvQ P = vQ vP

Also, the veloci ty of b ody Pre lat ive to b ody Q is given b y

vPQ = vP vQ

Clear ly their m agnitudes are equal but t heir d i rect ions are opposi te to each ot her.

Special cases of r elative velocity:

1 . Both the bodies are moving with the ve loc i ty in the samedirect ion, i.e. vQ = vP. In th is condi t ion ,

xQ P (t) = xQ (t) xP(t) = xQ (0 ) xP(0) = xQ P (0 )

I t show s tha t the d is tance be tw een the tw o bod ies does

not change with t ime. The re lat ive veloci ty is zero. The

posi t iontime graphs of the two bodies are stra ight l inesparal le l to each other as shown in f igure 25. I t is a lso clearf rom the graph tha t the d is tance be tween the two bod ies

rem ains constant at 10 m. The body P is behind body Q. The

body Pcannot catch t he body Q.

2 . The velocity of body Q is greater than the velocity of bodyP, i.e. vQ > vP. In th is cond i t ion , the d is tance be tw een the

two bodies continuously increases with t ime. The re lat ive


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veloci ty is vQP = vQ vP > 0. I t is posi t ive. The posi t iontime

graphs of the two bodies are stra ight l ines such that the

graph o f body Q has greater slope (velocity). The graphs are

shown in f igure 26. The body Pcannot catch the b ody Q.

3 . The velocity of body Q is smaller than the velocity of bodyP, i.e. vQ < vP. In th is condi t ion, the re lat ive veloci ty is vQP =

vQ vP < 0,i.e. vQ vP is negative. The graph of body P has

greater s lope than tha t o f the body Q. Initial ly (at t = 0), the

body P i s beh ind the body Q. Their in i t ia l posi t ions are 10 m

and 20 m , respectively. The body Povertakes the body Q atxQ (t) = xP ( t) = 30 m. Th is is the po in t a t wh ich they meet .

This happens at t = 15 sec. The graphs are shown in f igure

27 .

4 . The bodies P a nd Q are moving in opposite direct ions withvelocities vPa nd vQ . The veloci ty of bod y Q re lat ive to bo dy P

is given b y

vQ P = vQ + vP

The veloci ty of bod y Pre la t ive to body Q is given b y vPQ = vP + vQ

Clear ly their m agnitudes are equal but their d i rect ions are

oppo si te to each ot her. Their in i t ia l posi t ions are 10 m and 20

m, respectively. The body P is moving along + x-axis (+ ve

slope) and the bo dy Qalong x-axis ( ve slope). They m eet atxQ (t) = xP ( t) = 15 m. Th is is the po in t a t wh ich they meet

w hi le m oving in opposi te d irect ion s. This happens at t = 5 sec.

The graphs are show n in th e f igure 28.

Example 15A jet a i rp lane t rave l ling at the speed of 500 km / h e jects its

products of combust ion at the speed of 1500 km / h re lat ive to the jet p lane. W hat is the speed of the lat te r w i th respect t o an observer on

the ground?

Example 16Two t ra ins A and B of length 400 m each are moving on tw o

paral le l t racks w ith a uni form speed of 72 km/ h in the same direct ion,

w ith A a head o f B. The driver of B decides to overta ke A and

accelera tes by 1 m s2. I f aft er 50 s, the gua rd of B just brushes past t he

dr iver of A, what w as the or ig inal d istance betw een them ?

Example 17On a tw o- lane roa d, car A is t ravel l ing w ith a speed of 36 km/ h. Two cars B and C approach car A in

opposite d irect ions w ith a speed of 54 km / h each. At a certa in instant , w henthe dista nce AB is equa l to AC, both b eing 1 km , B decides to overt ake A before

C does. W hat m inimum accelerat ion of car B is required to avo id an accident?

Exam ples on graph s:

Example 18 Refer to the f igure 29 which shows the posi t ion-t ime graphs for

part icles P, Q, R, S and T. Answ er the follow ing questions.

1 . W hich pa rt icle / part icles is/ are m oving w ith a constant veloci ty?2 . W hich part icle / part icles has/ have non-zero a ccelerat ion?


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1

3 . W hich pa rt icle / part icles is /a re sta t ionary?4 . W hich particle/ part icles changes/ change its/ their direction?5 . W hich part icle has great est speed?6 . W hich part icle has low est non-zero speed?7 . W hich part icle has the grea test accelerat ion?

Example 19

Refer to the figure 30 which shows the velocity-time graphs forpart icles P, Q, R, S and T. Answ er the following questions.

1. W hich pa rt icle / part icles is/ are m oving w ith a constant veloci ty?2. W hich pa rt icle / part ic les has/ have non-zero

accelerat ion/retardat ion?

3. W hich pa rt icle / part icles is/ are stat ionary?4. W hich particle/ part icles changes/ change its/ their direction?5. W hich pa rticle has greate st velocity?6. W hich part icle has low est non-zero accelera tion?

Exam ples on p osit ion-t im e graphs:

1 . Sketch a position-time graph for a particle, moving with a constant,positive ve locity.Answer: Graph B of f igure 3.

2 . Sketch a position-time graph for a particle, moving with a constant,negative velocity.Answer: Graph C of f igure 3.

3 . Sketch a po sition-t ime gra ph for a part icle, moving in positive directionw ith a ccelerat ion.Answer: Graph D of f igure 3.

4 . Sketch a po sition-t ime gra ph for a part icle, moving in positive directionw i th re ta rda t ion. Answer: Graph OP of f igure 5.

5 . Sketch a position-time graph for a particle, moving in negativedirection with acceleration. Answer: Graph PQ of f igure 5.

6 . Sketch a position-time graph for a particle, moving in negativedirect ion w ith retardat ion. Answer: Figure 31 .

7 . Sketch a position-time graph for a particle moving in the +ve direction-init ially with a low constant speed and then a high constant speed.Answer: Figure 32 graph A.

8 . Sketch a position-time graph for a particle moving in the ve direction-init ially with a low constant speed and then a high constant speed.Answer: Figure 32 grap h B.

9 . Sketch a position-time graph for a particle moving in the +ve direction-init ially with a high constant speed and then a low constant speed.Answer: Figure 33 graph A.

10 .Sketch a position-time graph for a particle moving in the ve direction-in i t ia l ly wi th a h igh constant speed and then a low constant speed.

Answer: Figure 33 grap h B. 11 .Sketch a position-time graph for a particle moving in the +ve direction

at low constant speed and then in ve direction at high constant

speed.Answer: Figure 34 graph A.12 .Sketch a position-time graph for a particle moving in the +ve direction

at h igh constant speed and then in ve direction at low constant

speed.Answer: Figure 34 graph B.13 .Sketch a position-time graph for a particle moving in the ve direction


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1

at low constant speed and then in +ve direction at high constant speed.Answer: Figure 35 graph A.

14 .Sketch a position-time graph for a particle moving in the ve directionat high constant speed and then in +ve direction at low constant speed.Answer: Figure 35 grap h B.

Exam ples on velocity-t im e graphs:

1 . Sketch a velocity-time graph for a particle, moving with a constant,positive velocity. Answer: Graph A of f igure 9 or graph P of

f igure 29.2 . Sketch a velocity-time graph for a particle, moving with a constant,

negative velocity. Answer: Graph T of f igure 29 .3 . Sketch a velocity-time graph for a particle at rest. Answer: Point 3 of

f igure 29.4 . Sketch a velocity-time graph for a particle moving in the +ve direction,

accelerat ing f rom low speed to high speed. Answer: Graph B of f igure9.

5 . Sketch a velocity-time graph for a particle moving in the +ve direction,retarding from high speed to low speed. Answer: Graph C of f igure

9.6 . Sketch a velocity-time graph for a particle moving in the ve direction,

accelerat ing f rom low speed to high speed. Answer: Graph A of

f igure 36.7 . Sketch a velocity-time graph for a particle moving in the ve direction,

retarding from high speed to low speed. Answer: Graph B of f igure

36.8 . Sketch a velocity-time graph for a particle moving in the +ve direction-

init ially with a low constant speed and then a high constant speed.

Answer: Figure 37 graph A.9 . Sketch a velocity-time graph for a particle moving in the +ve direction-

init ially with a high constant speed and then a low constant speed.Answer: Figure 37 grap h B.

10 .Sketch a velocity-time graph for a particle moving in the ve direction-init ially with a low constant speed and then a high constant speed. Answer: Figure 38 graph A.

11 .Sketch a velocity-time graph for a particle moving in the ve direction-in i t ia l ly wi th a low constant speed and then a high constant speed. Answer: Figure 38 grap h B.

12 .Sketch a velocity-time graph for a particle moving in the +ve direction-in i t ia l ly wi th a constant speed and then with accelerat ion. Answer:Figure 39 graph A.

13 .Sketch a velocity-time graph for a particle moving in the +ve direction-in i t ia l ly wi th a constant speed and then with retardat ion. Answer:

Figure 39 graph B.

Additional Problems:


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1. A rif le bullet loses 1/ 20th of its velocity in passing through a plank. W hat is the lea st num ber of suchplanks required just to stop t he bullet?Answer:11

2 . Two ends of a train moving with constant acceleration pass a certain point with velocities v1 and v 2 ,the n w hat is the velocity of its m iddle point wh en it passes tha t point?Answer:

3 . Two trains are moving on the same track in the same direction with velocities v1 ( running ahead of thetwo) and v 2 such that v2 > v1 . I f a retardation a is produced in the second train, then what is the

minimum time to avoid coll ision?Answer:

4 . Two trains are moving on the same track in the same direction with velocities v1 ( running ahead of the

tw o by a d istance d) and v 2 such tha t v2 > v1 . I f a retard at ion a is produced in the second tra in, then what

is the m inimum value of d t o avoid collision?Answer:d =()

5 . A body moves from A to B w ith a consta nt speed of 20 m / s and returns f rom B to A w ith a consta nt

speed of 40 m / s. W hat is the average speed of the body?

Answ ers- Test your kno w ledge:1.

a . As th e size of a cr icket bal l is m uch smal ler than distance of the b oun dary from the keeper , the bal l can bet reated as a po in t ob ject .

b . As the size of a motor-bike is much smal ler than distance between Ghaziabad and Chandigarh, the bikecan be treat ed as a point ob ject.

c. As the size of a footbal l is not negl igible as compared to a distance of 4 m between two footbal l p layers,the footba l l cannot be t reated as a po in t ob ject .

d . The driver of the car can be considered as a point object because his size is much smal ler than thedistance covered by th e car.

e. A cup fal l ing off the edge of a table cannot be considered as a point o bject because the size of the cup isnot neg ligib le as com pared to the he ight o f t he tab le .

2. The d istance o f t he m an is equal to t o ta l path length w hich is 5 km + 5 km = 10 km but t he d isp lacement iszero, because th e ini t ia l and f inal po sit ion s are same.3. In half revolut ion , the displacement = diam eter o f the path = 2r and distance = th e circum ference = r

4. The d i s tance cove red by a pa r t i c l e i s equa l to the magn i tude o f d i sp lacemen t o f pa r t i c l e i n a un i fo rmm ot ion along a str aight l ine in a given direct ion.

5. Yes. Rest and mot ion are re la t ive terms. A body a t rest in one f rame of re ference may be in mot ion wi threspect t o another f rame of re ference. For example, a moving car on a road is a t rest w i th respect to t he dr iver

but in m ot ion w i th respect t o a man stand ing on the road.

6. A straight l ine, not p aral le l to t im e axis. A str aight l ine par al le l to t im e axis indicates zero veloci ty. 7. Slope of displacement -t im e (x-t) graph is equal to the veloci ty of t he body. 8. Given, x t 2 means v = dx/d t t and accelerat ion = constant. So, the body is moving with uniform

accelerat ion.

9. Parabola. Also see t he previo us answer. 10 . This is possible wh en the bo dy retu rns to the start ing poin t.11 . Speedo m eter m easures the instantaneous speed.12 . Yes, in the case of uniform circular motion, the magnitude of velocity remains constant but its direction changes

f rom po in t to po in t .

13 . I t is not possible as constant veloci t y means m agnitude (speed) and direct io n wi l l rem ain unchanged.14 . Str aight l ine p aral le l to t im e axis (Fig. 16)15 . Area o f ve locity- t im e graph = d isp lacement = 0 as the d isp lacement becomes zero w hen the body re tu rns to

the ground.

16 . Change in veloci ty b ecause v u = a t (= accelerat ion t im e = area)17 . The m ot ion o f t he body is re tarded.


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18 . I t is mo ving w ith u niform accelerat ion as distance travel led by i t in each successive second in creases by a f ixedvalue (= 4 m).

19 . Yes, i t is possible. At t he extrem e posit ion of an osci llat ing pendulum and at the h ighest p oint of a body throw nvertically upw ards, th e velocity is zero, but the acceleration is not zero.

20 . Yes, a body can tr avel oppo site t o it s acceleratio n. A body pr ojected upw ards wit h some initial velocity is subjectedto a dow nw ard acceleration (g).

21 . Accelerat ion is zero. The veloci ty-t im e graph w i l l be a str aight l ine paral le l to t im e-axis.22 . A m oving body w i l l have accelerated m otio n i f i ts: (a) direct ion of m otion changes (b) the magnitude of veloci tychanges; and (c) both d irect ion and m agnitude of veloci ty change.

23 . This is not p ossible. I f speed changes, the veloci ty w i l l change and hence the re w i l l be an accelerat ion .24 . Both of the bal ls wi l l h i t the ground with the same speed as their in i t ia l vert ical speeds and accelerat ions are

same.

25 . Both of the bal ls wi l l h i t the ground at the same t ime as their in i t ia l vert ical speeds (both = 0) andaccelerat ion s are same and t hey cover t he same vert ical distance.

26 .Solut ion s of exa m ples:

Exam ple 1As shown in the Fig. example 1, the length of AC is equal to the

m agni tude o f t he d isp lacement vector .So, AC = + = 5 kmAngle made by t he d isp lacement vector wit h respect t o East is given by

t an = BC/ AB = 4/3

Or = 530

(Nort h of East)Please note th at th e distance travel led is = 3 km + 4m = 7 km

Exam ple 2Distance travel led by dru nkard in 1 s = 1 m

Distance travel led by drunkard in 5 step s

= 5 1 m = 5 m

Tim e taken to t rave l 5 m = 5 sDistance travel led in 3 backward steps =

3 1 m = 3 mTim e taken to t rave l 3 m backw ard = 3 s

Net displacem ent in f i rst 8 s (5 s + 3 s) = 5

3 = 2 m

Net displacem ent in 16 s = 2 2 = 4 m

Net displacem ent in 32 s = 4 2 = 8 mIn the next 5 s he wi l l t rave l 5m .

His net displacement at the end of t hese5 s = 8 m + 5 m = 13 m

It m eans at t he end of t hese 5s, he wi l l fa l l into t he pi t .

So, tot al t im e taken t o reach th e pi t = 32 s + 5 s = 37 s.

Exam ple 3 Let the t o ta l t ime taken be 2 t .

Distance travel led in t im e t is d1 = v 1 t = 80t

Distance travel led in next half t im e t is d 2 = v 2 t = 40t


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Total distance = 60 km

Or 80t + 40t = 60

Or 100t = 60

Or t = 0 .60 hourTherefore, average speed v = tot al distance/ tot al t ime = 60/ (2 0.60)

= 60/1 .2 = 50 km/ h

Exam ple 4The veloci ty is given by

v = = 0 +2b t = 2 5. 5 t = 1 1 t

In i t ia l veloci ty, (at t = 0), v= 0 m/ s. The veloci ty at t = 4.0 s, v= 2b t = 2 5. 5 4 = 44 m / s.

Average veloci ty vav =(.) (.)

.. =

. = 55 m/ s.

Exam ple 5Given x = 12 t

2 4 t + 6

The veloci ty is given by

v= = 12 2t 4 + 0 = 24t 4

For in i t ia l veloci ty, t = 0Therefore, t he ini t ia l veloci ty is v = u = 24 0 4 = 4 m / s

Exam ple 6The slope of t he posit ion--t im e graph gives th e instant aneous veloci ty at a point. I t is clear from the graph

th at th e slope is posit ive betw een point s O and P but steadi ly decreases on th e way up. I t m eans the instant aneous

veloci ty continuously decreases from point O to point P. This is retarded motion. The slope becomes zero at the

top (point P), i .e. the instantaneous veloci ty is zero at point P (due to retardation). From point P to Q, the slope

becomes more and more negative. I t means the instantaneous veloci ty continuously increases from point P topo in t Q but in opposi te d i rect ion ( i .e. in the d i rect ion o f r e tardat ion) .

At point P, the slope changes from + ve to ve, i .e. the direct ion of motion gets reversed. As i t is a

parabol ic curve (x t2), the correspon ding veloci ty-t im e graph (See f igure 10) wi l l be a straight l ine (x t

2 v t ).

Exam ple 7 The slope of the veloci ty--t ime graph gives the instantaneous accelerat ion at a point. I t is clear from the

graph that the slope is posit ive between points O and P but steadi ly decreases on the way up. I t means the

instantaneous accelerat ion continuously decreases from point O to point P. The veloci ty increases but at adecreasing rate. This is accelerated motion up to point P. The slope becomes zero at the point P, i .e. the

instantaneous ve loci ty is maximum at po in t P.

From po in t P to Q, the slope becomes mor e and m ore negat ive . I t m eans the instantaneous acce lera t ion

cont inuously increases f rom po in t P to po in t Q but in opposi te d i rect ion. I t m eans the ve loci ty decreases f rom

poin t P to Q and becom es zero a t po in t Q.

Not e that at point P, the slope changes from + ve to ve (i .e. accelerat io n tur ns into retardation)

but the d i rect ion o f mot ion remains the same throughout (po in t O to Q). The corresponding acce lera t ion- t ime

graph w i l l be a straight l ine as show n in f igur e 12 i f i t is a parabol ic curve (v t2 a t ).

Exam ple 8 The train is moving with uniform veloci ty as i t takes double the t ime to cross double the distance ( length

of tr ain + length of bridge). Hence, its accelerat ion is zero.

Exam ple 9 Betw een 0 and 5 sec, the veloci t y is posit ive constant . Its value is given by the slope of t he graph. Therefore,

v0- 5 = = 2 m / s

Betw een 5 and 15 sec, the veloci t y is zero (zero slope).


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From v= u + g t, we get

0 = 20.5 9. 8 tOr t = 2.1 sec

Example 14 Using the equat ion o f m ot ion

v2

= u2

+ 2axOr 0 = (10)

2+ 2 a 3 0

Or a = 100/60 = 5 / 3 m / s2

Again using the same equat ion

0 = (30)2

+ 2ax

Or x = 900 / (2 5/3) = 270 m

Example 15Veloci ty o f t he je t re la t ive to the ground Vjg = 500 km/ h ( taken posi t ive)Veloci ty of th e comb ustion (gas) relat ive to the jet V cj = -1500 km / h ( i t i s opposi te to the je t ve loci ty)

Veloci ty of the combustion (gas) relat ive to the ground V cg = Vcj + Vjg = - 1500 + 500 = - 1000 km/ h (oppo site to jet

veloci ty)

Example 16 Veloci ty of B relat ive t o A. VBA = 72 72 = 0

Accelerat ion of B, a = 1 m/ s2

Time t = 50 s

Distance travel led by B relat ive to A = VBA t + at2

= 0 + 1 50 50 = 1250 m

Example 17 Speed of car A, VA = 36 km / h = 36 1000 (m )/3600 s = 10 m/ s

Speed of car B, VB = speed o f car C, VC = 54 km / h = 54 1000 (m )/3600 s = 15 m/ sSpeed o f B relat ive t o A. VBA (same direct ion) = 15 10 = 5 m/ s

Speed of C relat ive t o A. VCA (opposite d i rect ion) = 15 + 10 = 25 m/ s

Distance AB = AC = d = 1 km = 1000 m

To determ ine the t im e taken by car C to r each car A, used = ut + at2

= V CA t + 0 = 25t

Or 1000 = 25t

Or t = 1000/ 25 = 40 sCar B has to cover 1000 m in 40 s.

d = ut + at2

Or 1000 = VBA t + at2

Or 1000 = 5 40 + a 40 40

Or 1000 200 = 800a

Or a = 800/800 = 1 m / s2

Example 18 1. The part icles Q an d R. They have constant non- zero slope. The constant non- zero slope m eans the constant

veloci ty.2. Only the par t ic le Shas accelerat ion . To have accelerat ion the p art icle should have a variable veloci ty. In other

w ords, the slope o f the po si t ion- t ime graph should change f rom po in t to po in t .

3. The part icles Pan d Tare stat ionary. Their posit ion is not changing wit h t im e.4. No p art icle changes i ts direct ion. The direct ion changes only wh en th e veloci ty of a part icle changes from + ve

t o ve veloci ty or from ve to + ve ve loci ty. In o ther words, the s lope o f the posi t ion- t ime graph should

change i ts sign (from + ve slope t o ve slope or fr om ve slope t o + ve slope).5. The part icle Qhas greatest speed because of steepest slope.6. The part icle Rhas lowest speed because of least slope.


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7. Only the par t ic le Shas accelerat ion . To have accelerat ion the p art icle should have a variable veloci ty. In otherw ords, the slope o f the po si t ion- t ime graph should change f rom po in t to po in t .

Example 19 1. The part icles Pan d T. They have zero slope. Zero slope o n a v-t graph m eans zero accelerat ion and he nce, the

constant ve loci ty.

2. Particle R has non-zero accelerat ion, part icle S has non-zero retardation and part icle Q has non-zeroaccelerat ion ini t ia l ly. Non-zero slope on a v-t graph show s an accelerated m otio n.3. A v-t graph l ine along t im e-axis indicates zero veloci t y. But th ere is no such l ine. Hence, there is no stat ion ary

part icle.

4. Particles Ran d Schange their direct io n. The direct io n of a part icle changes when i t s veloci ty changes from a +ve to ve (part icle S) o r f rom ve to +ve (part icle R). Thus, the line on a v-t graph wi l l pass f rom t he + to the -reg ion o f the graph.

5. Particle P has the greatest veloci ty. I t has constant posit ive veloci ty and part icle T has constant negativeveloci ty. Other p art icles have variable veloci ty bu t non e exceeds the veloci ty of P.

6. Particle Rhas the smal lest accelerat ion. The slope of t he v-tgraph is the measure of t he accelerat ion.Solut ion s of Add i t ional Problem s:

Problem 1Let the t hickness of each plank be d and ini t ia l speed u.

Speed aft er crossing the f i rst plank is v = u u/ 20 = 19u/ 20Using the equat ion o f m ot ion

v2

= u2

+ 2ad

Or (19u/ 20)2

= u2

+2 ad

Or 2ad = 361u2/ 400 u

2= 39 u

2/ 40 0

Total thickness D covered by th e bul let w hen i t f inal ly stops wi l l be given by

0 = u2

+2 aD

Or 2aD= u2

Therefore , number o f p lanks = =

= 400/ 39 = 10.25Therefore , the bu l le t w i l l stop in the 11 t h plank.

Problem 2Let the length of the tr ain be L.

Using the equat ion o f m ot ion for t he length o f t he t ra in

v2

= u2

+ 2ax

Or = + 2 aLOr 2aL = -- - - - - - - - (1)W hen half tr ain passes the given point, x= L/ 2 a nd le t t he spee d be v, t hen

2aL/ 2 = v2

Or aL = v2 - - - - - - (2)

From equatio ns (1) and (2)

2 (v2

) = Or v= Problem 3 In i t ia l relat ive speed of t he f i rst train wit h respect to the second t rain, u = v2 v1

There w i l l be no col l ision w hen t he f inal relat ive speed becom es zero, i .e. v= 0

Using th e equation v= u + a t, we get0 = ( v2 v1) at


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Or t =

Th is is the requ ired m in imum t im e.

Problem 4 In i t ia l relat ive speed of t he f i rst train wit h respect to the second t rain, u = v2 v1

There w i l l be no col l ision w hen t he f inal relat ive speed becom es zero, i .e. v= 0

Using th e equation v2 = u2 + 2 ad, we get0 = ( v2 v1)

2 2ad

Or d =()

Problem 5 Let d be the distance between A and B, t 1 be the t im e taken to travel from A t o B and t2 be the t im e taken to t rave lfrom B to A. Hence, t 1 = d/20 and t 2 = d /40

Average speed =

=

=

= 26.7 m/ s

Motion in a Straight Line (Revised)

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