EXAMPLES/NOTES:

UNIFORM RECTILINEAR MOTION

1. Car A at a gasoline station stays there for 10 minutes after Car B passes at a constant speed of 40 miles per hour. How long will it take Car A to overtake Car B if it accelerates at 4 m/s²

Solution:

Ta = Tb

Sa = S10 + Sb

Consider B:

v= st

Tb= sv= Sb17.88m /s

Sb=17.88Tb

Consider A:

Sa=VoTa+12aT ²=0+

12 (4 ms2 ) (T a2 )=2Ta ²

For S10:

S10=vt=( 17.88ms ) (600 s )=10728m

Substitute:

Sa=S10+Sb

2T ²=10728+17.88T

Answer: T = 77.85 seconds

2. A stone is thrown up from the ground with a velocity of 300ft/s. How long must one wait before dropping a second stone from the top of 600ft tower if the two stones are to pass each other 200ft from the top of the tower?

Solution:

Consider 1

s=Vot−12>²

400 ft=( 300 fts ) ( t )−12 ( 32.2 fts2 ) (t 2 )

t=17.19 s ,1.45 s

Consider 2

s=Vot+ 12g t2

200=0+12 ( 32.2 fts2 )(t 2 )

t=±3.52 s

How long must one wait?

t = 17.19 seconds – 3.52 seconds

Answer: 13.67 seconds

VARIABLE ACCELERATION

1. The motion of a particle is given by s=2t 4−t 3

6+2 t 2 where s is in feet and t inseconds. Compute the

values of v and a whent=2 seconds.

s=2t 4−t 3

6+2 t 2

v=8 t3−12t2+4 t

a=24 t 2−t+4

@t=2 seconds

v=8 (2 )3−12

(2 )2+4 (2 ) ; v=70 ftsec

a=24 (2)2−(2 )+4 ; a=98ft

sec2

2. A particle moves in a straight line according to the law s=t3−40 t where s is in m and t in seconds.

a. When t=5 seconds, compute v.

b. Find the average velocity during the 3rd to 4th seconds.

c. When the particle comes to stop, what is its acceleration?

a. s=t3−40 t

v=3 t2−40

a=6 t

@t=5 seconds

v=3(5)2−40 ; v=35msec

b. vave=total distancetotal time

=−96−(−93)

4−3=−31

; vave=−3 msec

@3rd second

s= (3 )3−40 (3 ) ; s=−93m

@4 thsecond

s=(4)3−40 (4 ) ; s=−96m

c. if v=0

0=3 t2−40

t=3.65 seconds

a=6 t=6 (3.65) ; a=21.9m

sec2

3. The rectilinear motion of a given particle is given by s=v2−9 where s is in m and v is in msec

. When

t=0, s=0, and v=3msec

. Determine s− t, v−t , and a−t relations.

s=v2−9

dsdt

=2 v dvdt

v=2va

a=12

a=v−vot

12=v−(3)t

; v=12t+3

s=vo t+12a t2=(3 ) t+1

2 ( 12 ) t2 ; s=3t+14t 2

MOTION CURVES

1. A particle starting with an initial velocity of 60 ft /s has a rectilinear motion with the constant deceleration of 10 ft/s2 . Determine the velocity and displacement at t = 9 sec.

Solution:

For and :

2. An auto travelled 1800 ft in 40sec. The auto accelerates uniformly and decelerates uniformly at

6 ft/s2, starting from rest at A and coming to stop at B. Find the maximum speed in fps.

Solution:

120

)

V

A6x – 6(40 – x) = 06x – 0 = 6(40 – x)6x = 240 – 6xx = 20

3. An Auto starts from rest and reaches a speed of 60 ft/s in 15 sec. The acceleration increases uniformly from zero for the first 9 sec after which the acceleration reduces uniformly to zero in the next 6 sec. Compute for the displacement in this 15 sec interval.

Solution:

V:

For and :

PROJECTILE MOTION

1. A golf ball is fired from the top of a cliff 50 m high with a velocity of 10 m/s directed at 45⁰ to the horizontal. Find the range of the projectile.

Vo cosө = xt

= ranget

10cos45⁰ = ranget

For t:

y = Vo sinөt - 12

gt2

-50 = 10sin45⁰t - 12

( 9.81) t2

t1 = 3.99 s (checked)

t2 = -2.55 s

For range:

10cos45⁰ = ranget

10cos45⁰ = range3.99

Range = 28.21 m

2. In figure 9-6.10, a ball thrown down the incline strikes it at a distance s = 254.5 ft. If the ball rises to a maximum height h = 64.4 ft above the point of release. Compute its initial velocity and inclination ө.

Horizontal motion:

Vo cosө = xt

3√10

= x254.5

x=241.44 ft .

Vertical Motion:

y = Vo sinөt - 12

gt2

1√10

= y254.5

y=80.48 ft .

For maximum height:

H = 64.4 ft.

y=V 2−(V o sinө)2

−2(32.2)

64.4=−(V o sinө)2

−2(32.2)

(V o sinө)2 = (2)(32.2)(64.4)

V o sinө=√ 64.42

V o sinө=64.4

V o=64.4sin ө

For ө :

-80.48 = Vo sinөt - 12

gt2

-80.48 = Vo sinө(241.44V ocosө

) - 12

(32.2)(

241.442

V o2 cos2ө)

-80.48 = 241.44 tanө - (16.1)241.442

64.4sin ө

2

cos2ө

-80.48 = 241.44 tanө - (16.1)241.442

64.42 tan2ө

tanө=1.33 ; ө=¿ 53.06⁰

tanө=−0.27 ; ө=¿ -15.11⁰

therefore,

V o=64.4sin ө

= 64.4sin 53.06⁰

Vo = 80.57 ft/

3. Find the take-off velocity that is just enough to clear the gap.

Using

Using horizontal motion formula:

Vo cosө = xt

Vo cos30⁰ = 17.32t

t = 17.32

V ocos30⁰

Using vertical motion formula:

y = Vo sinөt - 12

gt2

-22.2= Vo sin30⁰t - 12

(32.2)t2

-22.2= Vo sin30⁰(17.32

V ocos30⁰) - 12

(32.2)(17.32

V ocos30⁰)2

Vo = 14.14 ft/s

4. How high is the hill?

Using horizontal motion formula:

Vo cosө = xt

100cos60⁰ = 500t

t = 500

100cos60⁰

t = 10 sec

Using vertical motion formula:

y = Vo sinөt - 12

gt2

y= 100sin60⁰(10)- 12

(32.2)(10)2

y = -743.97 ft.

KINETICS

1. Determine P that will give the body an acceleration of 6 ft/sec2 µ = 0.20.

ΣFx=ma

R=ma=Px−F

322∗63.22

=Px−F

(1)322∗63.22

= 45∗P−0.2∗N

ΣFy=0

W=N+Py

(2) 322=N+ 35∗P

Substituing 2 in 1 we get:

P = 722.17 lbs.

2. Determine the acceleration of the system and tension in the chord. µ = 0.30

Consider 200 N Block

FBD:

ΣFv=ma

R=ma=W−T

200g

∗a=200−T

Consider 100 N Block

FBD:

ΣFx=ma

R=ma=T−F

100g

∗a=T−0.3∗100

a = 5.56 m/sec2

T = 86.67 N

3. Find the acceleration of the system and tension in the block.

Consider 300 N Block

FBD:

ΣFv=ma

R=ma=T−W

300g

∗a=T−300

Consider 100 and 200 N Block

FBD:

ΣFx=ma

R=ma=Wx 1+Wx2−F1−F 2−T

300g

∗a=35∗100+ 3

5∗200−0.2∗4

5∗100−0.3∗4

5∗200−T

T = 208 N

a = -3 m/sec2

4.

1. What is the tension in the card?

2. Acceleration of the blocks

3. Velocity of B after 2 seconds?

Consider B Consider A

T T T

a a

W

R=ma1=T−W R=ma2=981−2T

A

B

196.2 N

981 N

196.2 N

981 N

196.29

a1=T−196.2 9819a2=981−2T

For a1: For a2:

s=V ot+12a1 t

2 s=V ot+

12a2t

2

5=12a1t

2 1 2.5=1

2a2 t

2 2

Solve 1 and 2 simultaneously:

2=a1a2

a1=2a2

Therefore:

T = 327N a1 = 6.54 m

s2a2=3.27

m

s2

5

aA

aC aB

Determine the tension and acceleration of each blocks.

Consider block A:

T1

aA R=ma A

150 N R=T1−W

1509a A=T 1−150 1

Consider block B:

T2

R=maB

R=W−T 2

aB1009aB=100−T 1 2

100 N

Consider block C:

T2

150 N

100 N50 N

aC R=maC

R=T2−W

50N509aC=T 2−50 3

For T1 and T2:

T1 2T2−T1=ma

2T2=T1

A:

1509a A=2T 2−150

SA B:

1009

(a A+aB )=100−T 2

150N

Original

C:

509

(a A+aB ' )=12−50

SA

SB

SB’

For SA: For SB’: For SB:

SA=V o t+12aA t

2 SB '=

12aB ' t

2 SB=12aB t

2

SA=12aA t

2

But SB=SB’+SA

SB=12aB ' t

2 +12aA t

2

12aB t 2 ¿

12t 2(aA+aB ')

aB=(a A+aB ')

Equation 2 will now be:

1009.81

(aA+aB ' )=100−T 2

100N50N

1

1509.81

a A=2T 2−150 T2 = 120

aA = 5.836 m

sec2

aB’ = -7.848 m

sec2

3

509.81

(aA−aB ' )=T 2−50

509.81

(aB '−aA )=T 2−50 T = 70 N

aA = -0.58 m

sec2

aB’ = 3.46 m

sec2

ASSIGNMENT NO. 1

9-3.6 How fast must an automobile of the previous problem move in the last 8 minutes to obtain an average speed of 35 mph?

From the previous problem:

s=vt

s1=(30mph ) (12min )=6miles

s2= (40mph ) (20min )=13.33miles

s3= (30mph) (8min )=6.67miles

sT=26miles ; t=40min

v= 26miles

40min×60 sec1min

=39mph=v o

a=v−vot

=35mph−39mph40min

=−9mph40min

=−30mph2

9-3.8 On a certain stretch of track, trains run at 60 mph. How far back of a stopped train should a warning torpedo be placed to signal an oncoming train? Assume that the brakes are applied at once and retard the train at the uniform rate of 4 fps2.

vo=60mph=88 fps

v2=vo2+2as

0=882+2 (4 ) s

s=468 ft .

9-3.10 A ship being launched slides down the ways with a constant acceleration. She takes 4 seconds to slide the first foot. How long will she take to slide down the ways if their length is 900 feet?

s=vo t+12a t2 ; vo=0

s=12a t2

1=12a(42) ; a=0.125 fps2

900=12

(0.125 ) t2

t=120 sec ¿2minutes

9-3.12 A stone is dropped down a well and 5 seconds later the sound of the splash is heard. If the velocity of sound is 1120 fps, what is the depth of the well?

s=vo t+12a t2

For the stone, For the sound,

s= (0 ) t+ 12(9.81) t 2 s=v t2

s=4.905 t 2 s=(341.376) t2

t 12= s4.905

t 2=s

341.376

t 1+ t2=5 sec

(√ s4.905

) + (√ s341.376

¿=5 sec

s=103.65meters

9-3.14. A train moving with constant acceleration travels 24 ft during the 10th sec of its motion and 18 ft during the 12th sec of its motion. Find its initial velocity.

Solution:

S1oth = 24 ft @ t = 9 sec to 10 sec

S12th = 18 ft @ t = 11 sec to 12 sec

@ S1oth:

s=vo t+12a t2

24=vo(9)+12(9) t2 ; 24=9 vo+

12a(81)

24=vo(10)+12a(10)2 ; 24=10v o+

12a(100)

S9-10 ¿ vo+12a(19)

24=vo+9.5a (eq. 1)

@ S12th:

s=vo t+12a t2

18=vo(11)+12(11)t 2 ; 18=11vo+

12a (121)

18=vo(12)+12a(12)2 ; 18=12vo+

12a(144 )

S11-12 ¿ vo+12a(23)

18=vo+11.5 a (eq. 2)

Using the eq. 1 & 2:

vo=52.5 fps ; a=−3 fp s2

9-3.16. An auto A is moving at 20 fps and accelerating at 5 fps2 to overtake an auto B which is 382 ft ahead. It auto B is moving at 60 fps and decelerating at 3 fps2, how soon will A pass B?

Solution:

@ Auto A: @ Auto B:

s=vo t+12a t2 s=vo t+

12a t2

s=(20) t+ 12(5) t 2 (eq. 2) s−384=(60) t+ 1

2(−3)t 2 (eq. 2)

Subtract eq. 2 from eq. 1:

384=−40t+4 t2

t=16 sec

9-3.18. The rectilinear motion of a particle is governed by the equation s = r sin ωt where r and ωare constants. Show that the acceleration is a = -ω2s.

Solution:

s=r sinωt

u=r ; du=0

v=sinωt ; dv=cosωt

vdu+ud v

v=ωr cosωt

u=ωr ; du=0

v=cosωt ; dv=−sinωt

vdu+udv

a=−ω2r sinωt

Since s=r sinωt

Therefore: a=−ω2 s

9-3.20 A ladder of length L moves with its ends in contact with a vertical wall and a horizontal floor. If a ladder starts from a vertical position and its lower end A moves along the floor with a constant velocity vA, show that the velocity of the upper end B is vB = – vA tan Ɵ where Ɵ is the angle between the ladder and the wall. What does the minus sign mean? Is it physically possible for the upper end B to remain in contact with the wall throughout the entire motion? Explain.

Solution:

g=√L2−X2

V b=dydt

=12×1Y

(−2 X ) dydt

But X=V a t+dxdt

=V a

Therefore: V b=−XY

V a=−V a tan

When Ө = 90°, V b=∞, which is impossible.

9-3.22 The velocity of a particle moving along the x-axis is defined by v=kx³ -4x² + 6x, where v is in fps, x is in feet, and k is a constant. If x = 1, compute the value of the acceleration when x = 2 feet.

Solution:

At x = 2 feet

v = (1)(2)³ - 4(2)² + 6(2) = 4fps

a=dvdt

=3x ² dvdt

−8 x dvdt

+6 dvdt

v=dvdt

a=3 x2 v−8 xv+6v

Substituting v = 4 fps.

a=(3 ) (2 )2 (4 )−8 (2 ) (4 )+6 (4 )

Answer: a=8 fps²

ASSIGNMENT NO. 2

Determine the acceleration of the 2 blocks after touching each other. Determine the time at which the 300N block will touch the 100N block.

R=ma100=W sinθ−f 100=W sinθ−μ N1

Wga100=W sinθ−μ N1 ; N1=100cos 30°

1009.81

a100=100sin 30 °−0.2¿¿ ; a100=3.206m

sec2

s=vo t+12a100 t

2

s=12(3.206) t2 → (1)

R=ma300=W sinθ−f 300=W sinθ−μ N2

Wga300=W sinθ−μ N2 ; N2=300cos 30°

3009.81

a300=300sin 30 °−0.1¿¿ ; a300=4.055m

sec2

s+1=vo t+12a300 t

2

s=12

(4.055 ) t2−1 → (2)

Equating (1) and (2)

3.2062

t 2=4.0552

t2−1

0.4245 t 2=1

t 2=2.356 ; t=1 .535 seconds

Acceleration after touch:

a t=a100+a300=3.206+4.055 ; a t=7 .261m

sec2

ASSIGNMENT NO. 4

1044. An elevator weighing 3220 lb starts from rest and acquired an upward velocity of 600 ft per min in a distance of 20 ft. If the acceleration is constant. What is the tension in the elevator cable?

Given:

W = 3220 lb Sol’n:

v = 600 ft/min = 10 ft/sec v2=2as

s = 20 ft (10 )2=2a (20 )

Req’d: T a=2.5 ft

sec2

Wga=T−W

322032.2

(2.5 )=T−3220

T=3470 lb

1045. A man weighing 161 lb is in an elevator moving upward with an acceleration of 8 ft per sec2.

(a) What pressure does he exert on the floor of the elevator? (b) What will the pressure be if the elevator is descending with the same acceleration?

Given: Sol’n:

v

W

T

Wman = 161 lb (a) Wga=T−W

a = 8 ft/sec2 16132.2

(8 )=T−161

Req’d: T=201 lb

(a) Pressure he exert (b)−Wg

a=T−W

(b) Pressure if the elevator −16132.2

(8 )=T−161

Descends with the same T=121 lb

acceleration

1046. The block in Fig. P-1046 reaches a velocity of 40 ft per sec in 100 ft, starting from rest. Compute the coefficient of kinetic friction between the block and the ground.

Given: Sol’n:

v = 40 ft/sec v2=2as

s = 100 ft 402=2a (100 )

Req’d: Coefficient of kinetic a=8 ft

sec2

Friction, µWga=P−μN

16132.2

(8 )=60−μ (161 )

μ=0.124

1047. Determine the force P that will give the body in Fig. P-1047 an acceleration of 6 ft per sec2. The coefficient of kinetic friction is 0.20.

P= 60 lb

161 lb

3

4

P

322 lb

Given: Sol’n:

a = 6 ft/sec2 Wga=Px−μN

µ = 0.232232.2

(6 )=45P−(0.2 ) N

Req’d: force,P 0.2N=45P+60

P y+N−W=0

N=322−35P

32232.2

(6 )=45P−(0.2 )(322−35 P)

P=135.22

1053. Referring to Fig. P-1052, assume A weighs 200lb and B weighs 100lb. Determine the acceleration of the bodies if the coefficient of kinetics friction is 0.10 between the cable and the fixed drum.

Fig. P-1052

B A

Given:

W A=200 lb μ=0.10

W B=100 lb

Solution:

200−T A=20032.2

a 1

T B−100=10032.2

a 2

T A

T B

=eμθ

T A

T B

=e0.1(π )

T A=1.37T B 3

Substitute 3 to 1:

200−1.37T B=20032.2

a 4

From 2:

T B−100=10032.2

a

Substitute 2 to 4:

200−1.37 (100+3.11a )=6.21a

200−1.37−4.261a=6.21a

a=6.02 ft

sec2

1055. If the pulleys in Fig. P-1055 are weightless and frictionless, find the acceleration of the body A.

A

B.Fig. P-1055

For A:

200−T=ma

aA 200−T= 20032.2

aA

200 lb T=200− 20032.2

aA 1

For B:

T T 2T−300= 30032.2

aB

aB 300+ 30032.2

(aA2

)=2T 2

200 lb

300 lb

300 lb

Equate 1 and 2:

200− 20032.2

aA=300+ 300

32.2(aA2

)

2

a A=5 .85ft

sec2

1057. The coefficient of kinetic friction under block A in Fig. P-1057 is 0.30 and under block B it is 0.20. Find the acceleration of the system and the tension in each cord.

B

A

30o

At C,

300 – T2 = 30032.2

a -----1

At B,

T2 – T1 – 200sin30o – 200cos30o (0.2) = 20032.2

a

T2 – T1 – 134.64 = 6.21a -----2

200lb

100lb 300lb

At A,

T1 – 100sin30o – 100cos30o (0.3) = 10032.2

a

T1 – 75.98 = 3.11a -----3

T1 = 75.98 + 3.11a

Substitute T1 to 2,

T2 – (75.98 + 3.11a) – 134.64 = 6.21a

T2 – 210.62 = 9.32a

T2 = 210.62 + 9.32a -----4

Substitute 4 to 1

300 – 210.62 + 9.32a = 30032.2

a

89.38 = 18.64a

a= 4.8 ft/sec2 ans.

T1 = 75.98 + 3.11 (4.8) = 90.91 lb ans.

T2 = 210.62 + 9.32 (4.8) = 255.36 lb ans.

1059. Compute the acceleration of body B and the tension in the cord supporting body A in Fig. P-1059.

200lb

300lb

fh = 0.20 A

34

In block A,

200 – T = 20032.2

aA

In block B,

2T - 35

(300) - 45

(300) (0.20) = 30032.2

aB

2T – 228 = 30032.2

aB

In getting the acceleration for B,

Since aA = 2 aB

2 [ 200 – T = 20032.2

aA ]

+ -228 + 2T = 30032.2

(0.5) aA

400 – 228 = 550aA

172 = 500 (2aB)

aB / 32.2 = 172/1100

aB = 5.03 ft/sec2 ans.

1061. Compute the time required for the 100-lb body in Fig. P-1061 to move 10 ft starting from rest.

fh=0.20

3 4

For 100-lb block,

T1 - 35

(100) = 10032.2

a1

For 800-lb block,

45

(80) - 35

(80) (0.20) – T2 = 80032.2

a2

2T2 – T1 = ma

2T2 – T1 = 0

2T2 = T1

Since a2 = 2 a1

a2 = 2.82 ft/sec2

Solving for t1 (100lb),

100 lb

80lb

S = 12

a1t2

10 = 12

(2.82) t2

t = 2.663 sec. ans.

1063. Determine the acceleration of each weight in Fig. P-1063, assuming the pulleys to be weightless and frictionless.

A C

B B

For A,

T – 150 = 15032.2

aA

For B,

2T - 480 = 48032.2

aB

For C,

150 lb

480 lb

300 lb

300 – T = 30032.2

aC

Since aB= 12

aC - 12

aA

For Tension,

15032.2

aA = T -150

(480) (12

aC - 12

aA) = 2T - 480

30032.2

aC = 300 – T

T = 218.7 lb

Solving for acceleration,

218.7 – 150 = 15032.2

aA ; aA = 14.7476 ft/sec2 ans.

300 – 218.7 = 30032.2

aC ; aC = 8.7262 ft/sec2 ans.

2(218.7) – 480 = 480/32.2aB

aB = -2.85775 ft/sec2 or aB = 2.85775 ft/sec2 (downward) ans.

1065. Determine the maximum and minimum weights of the body C on Illustration Problem 1043 that will keep C stationary. All other data remain unchanged.

Solution:

N =gWF = 16W

N = 800

F = 160

1000

T

B

2T

800

aB = 12

WA

2B

800

W

F

8W

6W

ΣFx=wga

For A;

600 – 160 – T = 1000g

aA

For B;

2T – 800 = 800g

(12aA)

Solving for T;

T = 407 lb

For up plane impending motion of C:

ΣFx=0

T = 407 = 6W + 16W

W = 535 lb

For down plane impending motion of C;

ΣFx=0

T = 407 = 6W - 16W

W = 924 lb

1067. In the system of connected blocks in Fig 1067, the coefficient of kinetic friction is 0.20 under bodies B and C. determine the acceleration of each body and the tension in the cord.

Solution:

B

fh = 0.20 A

34

Direction of motion: Assuming at rest

T= 400

On B, Net force = 2T = 800 – 480 > F

= 128 (B rises)

On B, Net force = 600 – (T=400) > F

= 160 ( C falls)

With C at rest, Sa’ = 2Sb

With B at rest Sa “ = Sc

Net motion = Sa’ - Sa” = Sa = 2Sb - Sc

Differentiating :

aA = 2aB – aC

aB = 12(aA+aC)

C

fh = 0.20 3400lb

800lb

1000lb

4

ΣFx=wga

For ;

400 – T = 400g

aA

For B;

2T – 480 – 128 = 1000g

aC

Solving;

T = 348.2lb

aA = 4.18 fps2

aB = 3.57 fps2

aC = 2.96 fps2

1069. Two blocks A and B each weighing 96.6 lb and connected by a rigid bar of negligible weight move along the smooth surfaces shown in Fig 1069. They start from rest at the given position. Determine the acceleration of B at this instant. Hint: To relate aA to aB, use the method developed in Illus Prob on 258.

Solution:

V = dsdt,

LxVa+YVb=0

a = dvdt

Va + xaA + Vb + Yab = 0

At start,

Va = Vb = 0

aA = − yxab

or if Ab is down + down

aA = yxab =

86ab

ΣFx=wga

For A;

6P = 96.632.2

86ab= 4ab

For B;

96.6 – 8P = 96.632.2

ab = 3ab

Solving;

aB = 11.6 fps2

aA = 15.47 fps2

1071. The pulleys in the preceding problem have been assumed to be frictionless and weightless. What changes would there be in the solutions of these problems if the pulleys (a) had friction (b) had appreciable weight?

Solution:

(a) with friction, the tensions on the opposite sides of the pulley would be unequal.

(b) With appropriate weight, the supporting tension would not equal twice the outside tensions.

SEATWORK

A ball is dropped from the tower of 80 ft. high at the same instant that a second ball is thrown upward from the ground with an initial velocity of 40 ft/sec. When and where do they pass, and with what velocities?

SOLUTION:

h=v0 t+12g t 2

h=40 t−12

(32.2 ) t2(1)

80−h=12

(32.2 ) t 2

h=80−16.1 t 2(2)

Equate (2) and (1):

80−16.1t 2=40 t−12

(32.2 )t 2

t=2 sec .

Substitute t to (1) and (2):

h=80−16.1 (2 )2=15.6meters¿ the bottom

s'=80−15 .6=64 .4mete rs¿ the top

v f 2−vo2=at

v f 2=40−32.2 (2 )

v f 2=−24.4 ftsec

v f 1=32.2 (2 )

v f 1=64.4ftsec

v=v f 1+v f 2

v=64.4−24.4

v=40 ftsec

An automobile starting from rest speeds up to 40 ft/sec with a constant acceleration of 4 ft/sec2 run at this speed for a time and finally comes to rest with a deceleration of 5 ft/sec2. If the total distance travelled is 1000 ft, find the total time required.

SOLUTION:

v−v o=a t 1

40=4 t1

t 1=10 sec .

s1=12

(4 ) (10 )2=200 ft .

v=s2t 2

s2=40 t2

v−v o=−a t 3

−40=−5 t 3

t 3=8 sec.

s3= (40 ) (8 )−12

(5 ) (8 )2

s3=160 ft .

s1+s2+s3=1000

200+s2+160=1000

s2=640 ft .

640=40 t 2

t 2=16 sec .

T t=t 1+t 2+t 3=10+16+8

T t=34 sec .

The velocity of a particle moving along the x-axis is defined by v=kx3-4x2+6x where v is in m/s and x = meter and k = 1. Compute the acceleration when x = 2m.

SOLUTION:

When k =1 ;

v=(1 ) x3−4 x2+6 x

v=x3−4 x2+6 x

ads=vdv

a=v dvds

a=(x3−4 x2+6x ) (2x2−8 x+6)

When x = 2 m

a=[(2 )3− (4 )2+6 (2)][2 (2 )2−8(2)+6]

a=[2 ] [4 ]

a=8m /s2

a=6√v ; when t = 2 sec; v = 36 m/sec; s = 30 m. Determine s at t = 3 sec.

SOLUTION:

a=dvdt

6√v=dvdt

dv

√v=6dt

v−12 dv=6dt

∫ v−12 dv=∫6dt

v12

12

=6 t+c

2v12=6 t+c

When v = 36 m/sec; t = 2 sec

2(36)12=6 (2 )+c

c=0

Therefore,

2v12=6 t

Or

v=18 t2

v=dsdt

dsdt

=18 t 2

ds=18 t 2dt

∫ ds=¿∫18 t 2dt ¿

s=18 t3

3+c

s=6 t3+c

When s = 30 m; t = 2 sec

30=6 (2 )3+c

c=−18

Therefore,

s=6 t3−18

When t = 3 sec

s=6 (3 )3−18

s=144meters