chapter 2 speed velocity and acceleration

8/6/2019 Chapter 2 Speed Velocity and Acceleration

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Speed, Velocity and Acceleration



� 2. Kinematics

� 2.1Speed, velocity and acceleration

Candidates should be able to: ± state what is meant by speed and velocity ± state what is meant by uniform acceleration and

calculate the value of an acceleration using change invelocity/time taken ± interpret given examples of non-uniform acceleration



� 2.2 Graphical analysis of motion including Free-fall

± *plot and *interpret a distance-time graph and a speed-timegraph ± *deduce from the shape of a distance-time graph when a body

is:� ( i) at rest� ( ii) moving with uniform speed

� ( iii) moving with non-uniform speed ± *deduce from the shape of a speed-time graph when a body is:� ( i) at rest� ( ii) moving with uniform speed� ( iii) moving with uniform acceleration� ( iv) moving with non-uniform acceleration

± *calculate the area under a speed-time graph to determine thedistance travelled for motion with uniform speed or uniformacceleration

± state that the acceleration of free fall for a body near to the Earthis constant and is approximately 10 m/s 2



SPEED

Speed is the rate of change of distancewith time .

� T he SI Unit is m/sIt is commonly represented as km/h for transportation and km/s for the speed of light . ± Eg the speed of light is 3 00 000 km/s ± Eg the maximum speed of a truck is 6 0 km/h



ACCELERA TION

� T he rate of change of velocity .Or the rate of change of speed in a

specified directon

final velocity initial velocity

time taken togo from initialvelocity to

final velocity

a = (v ± u) ÷ t

acceleration



ACCELERA TION

What is the acceleration of a car thatincreased its speed from 10 m/s to 3 0 m/sin 4 seconds?

Ans: a = (30 m/s ± 10 m/s) ÷ 4s= 20 m/s ÷ 4s

= 5 m/s2



ACCELERA TION

What if it were decelerating (ie, slow down,go slower, decrease speed etc)?

Eg: the same car now slows down back to10 m/s in 5 seconds . What is hisacceleration?

Ans: a = (10 m/s ± 3 0 m/s) ÷ 5s= (- 20 m/s) ÷ 5s= - 4 m/s 2

Means slowing down



0

displacement

� 2.2 Graphical analysis of motion including Free-fall

± *plot and *interpret a distance-time graph and a speed-timegraph ± *deduce from the shape of a distance-time graph when a body

is:� ( i) at rest� ( ii) moving with uniform speed

Time0



(iii) moving with non-uniform speed

Graphs

Displacement

0Increasing speed

maximum speed

minimum speed

(at rest)

Time



*deduce from the shape of a speed-time graphwhen a body is:

(i) at rest(ii) moving with uniform speed(iii) moving with uniform acceleration(iv) moving with non-uniform acceleration

Graphical Analysis



Graphical Representations

Velocity

(i)

Time

(ii)

(iii) (iv)

(i) at rest(ii) moving with uniform speed(iii) moving with uniform acceleration(iv) moving with non-uniformacceleration



Other skills needed

You need to be able to:

± calculate the area under a speed-time graphto determine the distance travelled for motionwith uniform speed or uniform acceleration

± state that the acceleration of free fall for a

body near to the Earth is constant and isapproximately 10 m/s 2



Area Under A Graph

Velocity /m/s

(i)

(ii)

(iii)

Time/s0 10 1 4 23

A bus stopped at a bus-stop for 10 seconds before accelerating to avelocity of 15 m/s in 4 seconds and thenat a constant speed for the next 9

seconds . How does the graph look like?

15



Area Under A Graph

Velocity /m/s

(i)

(ii)

(iii)

Time/s0 10 1 4 23


seconds .

15



Area Under A Graph

Velocity /m/s

(i)

(ii)

(iii)

Time/s0 10 1 4 23


seconds .

15 How far did thebus go in this 23seconds?



Let¶s calculate!

Distance travelled in first 10 seconds iszero!

Distane travelled in the next 4 seconds is½ x 4 x 15 = 3 0 mDistance travelled in the final 9 seconds issimply 9 x 15 = 135 m

� T otal distance travelled = 165 m



velocity /ms -1 0 10 20 30 40 50 60

time / s 0 1 2 3 4 5 6

6050403020

10

0

velocity/ms -1

0 1 2 3 4 5 6 7

time/s

a cceler a tion = µgr ad ient¶ = (60-10)/(6-1)= 10 ms -2

Ex a mplePlot the velocity-time gr a ph of a moving object. Hence, fin d itsconst a nt a cceler a tion.

Author - C HIJSNGS



A bullet tr a in st a rts from rest from a st a tiona n d tr a vels a long a str a ight horizont a l tr a ck tow a r d s a nother st a tion. The gr a ph in fig.shows how the spee d of the tr a in v a ries withtime over the whole journey. Determine:(a ) the tot a l d ist a nce covere d by the tr a in,(b) the a ver a ge spee d of the tr a in.

Spee d / ms -1

Time / s

40

0 2 12 16

(continue on next slide)

Author - C HIJSNGS



Solution

Tot a l d ist a nce tr a velle d

= µa re a un d er the gr a ph¶= (1/2)(10 + 16)(40)= 520 m

A ver a ge spee d = (tot a l d ist a nce) / (tot a l time )= 520 / 16

= 32.5 ms -1

Author - C HIJSNGS



The m a ximum possible a cceler a tion of a ca r is2.5 ms -2. Its m a ximum d eceler a tion is 5 ms -2, a n d

its gre a test spee d is 50 ms -1. Fin d the shortertime require d

for the c a r totr a vel 2.1 kmfrom rest torest.

(continue onnext sli d e)

v/ms -1

t/ss1 s2 s3

50

0 t1 t2 t3

Author - C HIJSNGS



Since a = µgr ad ient¶ = v/ttherefore, 2.5 = 50 / t 1 t1 = 20 s

5 = 50 / t 3 t3 = 10 s

a lso s = µ a re a un d er the gr a ph¶we h a ve, s 1 = (1/2) x 50 x 20 = 500 m

s3 = (I/2) x 50 x 10 = 250 m

therefore d ist a nce tr a velle d a t uniform spee d

= 2100 - 500 - 250= 1350 m

hence, t 2 = 1350 / 50 = 27 sso,

tot a l time t a ken = 20 + 27 + 10 = 57 s

Solution

Author - C HIJSNGS



Acceleration Of Free FallVelocity /m/s

Time /s

Objectthrownupwards witha startingvelocity of 5m/s

Gradient of line is 10 m/s 2

Objectthrowndownwardswith astartingvelocity of 5m/s .

Gradient of line is 10 m/s 2

5

0

chapter 2 speed velocity and acceleration

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