Velocity & Acc_Engineering Science DMV1032_2132

8/6/2019 Velocity & Acc_Engineering Science DMV1032_2132

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Engineering Science:

Chapter 2 ± Velocity & Acceleration



Velocity & Acceleration

1. As a first step in studying classical mechanics, we

describe motion in terms of spaces and time while

ignoring the agents that caused that motion.

2. This portion of classical mechanics is called

kinematics.3. In this chapter we consider only motion in one

dimension, that is, motion along a straight line.

4. We first define position, displacement, velocity

and acceleration.

5. Then, using these concepts, we study the motion

of objects traveling in one dimension with a

constant acceleration



2.1 Displacement

A displacement is the shortest distance from the initial and finalpositions of a point P. Thus, it is the length of an imaginarystraight path, typically distinct from the path actually traveled byP.

A displacement vector represents the length and direction of

that imaginary straight path.

A

B

P



2.1 Displacement

When we say that a person has walked 7 kilometers, we give noindication either of the actual distance between his initial and finalpositions or of the direction of the final position relative to theinitial position.

He could be at any point within a radius of 7 km: he could even

be back at his initial position. Distance is therefore a scalar quantity, i.e it has magnitude

only ( no direction).

7km 3km

4km



2.1 Displacement

A person has walked 3 km eastwards, as represented by AB in fig.

and then walked 4 km northwards, as represented by BC,

his final position C is 5 km away from his initial position A.

This change of position is termed the displacement and is

independent of the path followed and of the time taken; thus, hecould have reached C by going north-east from A to D and thensouth-east from D to C, as shown by the dotted lines in figurebelow.

3 km A B

C

D

4 km



2.1 Displacement

Since displacement has both magnitudeand direction, it is a v ector quantity andcan be represented by a straight line drawnto scale in the direction of thedisplacement.

From figure, it will be seen that during thefirst part of the 7 km journey from A to C,the displacement is 3 km in an easterly

direction and this can be represented by ahorizontal vector AB, drawn to a scale of,say, 1 cm to 1 km. During the second partof the journey, the displacement is 4 km ina northerly direction and is represented infig. 6.1 by vector BC, 4 cm long, drawnvertically at B.

The resultant displacement is represented

to scale by the straight line AC. By measurement, the length of AC is found

to be 5 cm and the angle between AC and AB is 53o;

i.e. the resultant displacement is 5 km ina direction 53o north of east and can bedetermined by adding vectorially thecomponent displacements

3 km A B

C

D

4 km

53o



2.2 Speed

If a motor car travels 1.6 km in 2 minutes, its speed is 0.8kilometers per minute, or

0.8 km/minutes x 60 minutes/hour = 48 km/hr

In general, we can say that the speed of a body is the

distance traversed in unit time, or the rate at whichdistance is traversed .

The distance can be expressed in any convenient unit suchas a meter, a kilometers, etc.,

and the unit of time can also be any convenient value suchas an hour, a minute or a second .

1.6km

t = 2 minutes

V = Distance, s

time, t

= 1.6 km

2 min

= 0.8km/min



2.2 Speed

If a motor car travels a distance of 48 km in one hour, its av eragespeed is 48 km/h, but it is extremely unlikely that the car willtravel at exactly this speed during the whole hour ± its speed willbe at times higher and at other times lower than this value.

A body has constant speed only if it moves over equal distances

in equal intervals of time ± however short the interval. The average speed of a body is the total distance divided by the

time; thus, if a body travels a distance s meters in t seconds, theaverage speed, v meters per second, is given by:

V = s [meters] = s meters/second

t [seconds] t

V = s [meters] = s meters/second

t [seconds] t



2.2 Speed

Example:I f a motor car is traveling at a speed of 100 km/h,

what is the speed in meters/second?

I f a motor car is traveling at a speed of 100 km/h,

what is the speed in meters/second?

100 km = (100 x 1000) meter

1 hour (1 x 3600) seconds

= 100,000 meter

3600 seconds

= 27.78 m/s

100 km = (100 x 1000) meter

1 hour (1 x 3600) seconds

= 100,000 meter

3600 seconds

= 27.78 m/s

Note:

1km = 1000m

1hr = 60min = 3600s



2.2 Speed

Example:If an aeroplane travels a distance of 800 km at a constant

speed in 12 h, calculate:

a) its speed in m/s

b) the number of km traveled in 20 minutes, and

c) the time taken to travel 100 km.

If an aeroplane travels a distance of 800 km at a constant

speed in 12 h, calculate:

a) its speed in m/s

b) the number of km traveled in 20 minutes, and

c) the time taken to travel 100 km.

s = 800km

= 800 x 1000

= 8 x 105m

t = 12 hrs

= 12 x 3600= 43200s

a) v = s/t

= 8 x 105m

43200s

= 18.52m/s

b) t = 20min = 20 x 60 = 1200s

v = s/t

s = v x t

= 18.52 x 1200

= 22224m

= 22.2km

c) s = 100km = 100 x 1000

= 1 x 105m

v = s/t

t = s/v

= 1 x 10518.52

= 5399.57s

= 1.5h



2.3 Graph Relating Distance, Time

& Speed

Distance vs time & speed vstime graph for constantspeed.

Constant speed - no

speed change The slope at graph (a)

represent speed at whichthe body is moving.

The area shown shaded inspeed vs time:

= speed x time

= 5m/s x 20s

= 100m




& Speed

Distance vs time & speed vs

time graph for varying speed.

Average Speed:

= 100m/20s = 5m/s

The slope at initial and final

are zero (graph a).

Slope at any intermediate is

obtained by drawing a tangent

line.

Tangent line




& Speed Slope AB:

4.5s 5.5s, average speed = 5.6m/s

So, distance:

s = 5.6m/s x 1s

= 5.6m

5.6m, is represented by the area of the shade trip.

Total area below the graph (b)represent the total distance traveled.

Tangent line

m/s5.6

2.5)s-(10

42m

ABof Slope

!

!

!

AC

BC

5.6 m/s

5.5s-4.5s = 1s




& Speed

t, is the length of the time taken

Distance travel = area enclosed by graph

= average speed x time

= vt

6

vvvvvv vSpeedAverage

654321 !!

v1

v2

v3 v4

v5

v6



2.4 Linear Velocity

Speed is scalar quantity

Speed of a body can be stated without any

reference to the direction.

Velocity is vector quantity Velocity has both magnitude and direction.

If a car traveling in northerly direction at a speed

40km/h, the velocity is said to be 40km/h

northwards.



2.5 Relative Velocity

B

A v = 50km/h

v = 80km/h

Train A is moving at: 80 ± (-50) = 130km/h, relative to train B.

Observer at B, train A would appear to be eastward ()traveling at 130km/h.

Observer at A, Train B would appear to be travelingwestward () at 130km/h.



2.6 Acceleration

Velocity increase - accelerating

Velocity decreasing - retarding

Acceleration = Rate of change of velocity

The car velocity increase 1.5m/s every second from 0m/s to 30m/s

0m/s ? m/s ? m/s ? m/s 30m/s



2.6 Acceleration

u = initial velocityv = final velocity

a = acceleration

Suppose the velocity to increase at

a uniform rate to v in time,t

0m/s ? m/s ? m/s ? m/s 30m/s

at uv

t

uv

aon Accelerati

uv

!

!

!

!!

!

time

velocityoChange

velocityochangeoRate

VelocityoChange

Unit for acceleration:

m/s

2



Substituting the value v from equation

(1) to equation (2);

at uvt

uv

aon Acceler ati

uv

!

!

!

!!

!

time

velocityof hange

velocityof changeof ate

elocityof hange

2.6 Acceleration

t vu s

s

vu

s

!

!

!

!

21

x timevelocityaverage

21 velocityaverage

travelDistance

1

2

2

21

2

1

at ut s

t at uu s

!

!

3



2.6 Acceleration

a suv

at ut au

at uat u

at uvif

at uv

2

)21(2

)(2

)(,

22

22

22

22

!

!

!

!

! 1

4Note= a is positive when accelerating

a is negative when retarding



Summary

asuv

at ut sceTravel Dis

at uvt

uvaon Accelerati

t

svSpeed

22

1;tan

;

;

22

2

!

!

!

!

!



Thank You

Velocity & Acc_Engineering Science DMV1032_2132

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