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
Answers
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.