Question

A car starts from rest and moves with constant acceleration of 4 m/s2 for 30 seconds. Then the braattained by car (ii) Magnitude of retardationkes are applied and the car comes to rest in another 60 seconds. Find [5] (i) Maximum velocity of car (iii) Total distance covered by car

Answer 1

Explanation:

Train starts from rest, hence the initial velocity u = 0.

It moves with acceleration = 2m/s2 for half minute (30 seconds).

Distance covered in this time interval is given by:

S=ut+½at

2

=0+½×2×30×30

=900m

Velocity attained by this acceleration after 30 seconds:

v=u+at

=>v=0+2x30

=>v=60m/s

From this velocity, brakes are applied and train comes to rest in 60 seconds.

The retardation is given by:

v=u–at

=>0=60–a×60

=>a=1m/s

2

Distance covered in this time:

$$V2= u2 + 2aS$$

=>0=(60)2+2(−1)S

=>0=3600–2S

=>S=3600/2=1800m.

So, total distance moved =900m+1800m=2700m.

Maximum speed of the train=60m/s.

Position of the train at half its maximum speed.

Here, you need to note that first the train is accelerating to 60 m/s, and then it is decelerating to 0 m/s. So there are two positions when speed is 30 m/s.

(I) When the train is accelerating with an acceleration of 2 m/s,

time at which speed = 30m/s is:

v=u+at

=>30=0+2xt

=>t=15s

At 15s, distance covered from origin is:

S=ut+½at 2

=0+½×2×15×15

=225m

(II) When the train is retarding from 60 m/s to 0 m/s, at a retardation of 1 m/s2 , time at which the speed reaches 30 m/s is:

v=u–at

=>30=60–1xt

=>t=30s

At 30s, distance covered is:

S=ut–½at 2

=60x30–½x1x(30)2

=1800–(15x30)

=1800–450

=1350m (from the initial 900m covered).

So, distance from origin =900+1350m=2250m.