A car starts from rest and moves with constant acceleration of 4 m/s2 for 30 seconds. Then the brakes are applied and the car comes to rest in another 60 seconds. Find [5] (i) Maximum velocity attained by car (ii) Magnitude of retardation of car (iii) Total distance covered by car (iv) Average speed of car through out the motion
Answers
Answer:
A train starts from rest and moves with a constant acceleration of 2.0 m/s
2
for half a minute. the brakes are then applied and the train comes to rest in one minute. Find
the total distance moved by the train,
January 17, 2020avatar
Sandeep Kosvi
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ANSWER
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.
Explanation:
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Given:
The initial velocity of the car, u = 0 m/s (at rest)
The acceleration of the car, a = 4 m/s²
The time of acceleration, t₁ = 30 seconds
The time of retardation, t₂ = 60 seconds
To Find:
(i) Maximum velocity attained by car.
(ii) Magnitude of retardation of car.
(iii) Total distance covered by car.
(iv) Avg speed of the car through out the motion.
Calculation:
(i) Maximum velocity attained by the car can be given by the 1st equation of motion as follows:
v = u + at₁
⇒ v = 0 + 4 × 30
⇒ v = 120 m/s
(ii) For retardation of the car (a'):
Initial velocity will be = v = 120 m/s
Final velocity will be, v' = 0 m/s
⇒ v' = v + a' × t₂
⇒ 0 = 120 + a' × 60
⇒ a' = - 2 m/s²
(iii) Total distance travelled by the car during whole motion = Distance travelled during acceleration + Distance travelled during retardation
⇒ S = s₁ + s₂
⇒ S = (u × t₁ + 1/2 × a × t₁²) + (v × t₂ + 1/2 × a' × t₂²)
⇒ S = (0 + 1/2 × 4 × 30²) + (120 × 60 + 1/2 × (-2) × 60²)
⇒ S = 1800 + (7200 - 3600)
⇒ S = 1800 + 3600
⇒ S = 5400 m
(iv) Avg speed of the car = Total distance travelled by the car / Total time taken
⇒ Avg Speed = 5400/90
⇒ Avg Speed = 60 m/s
- So, the maximum velocity attained by the car is 120 m/s; the magnitude of retardation of the car is -2 m/s²; the total distance covered by the car is 5400 m and the average speed of the car through out the motion is 60 m/s.