A train starting from rest attains a velocity of 72 km/h in 5 minutes. Assuming that the acceleration is uniform, find the acceleration and the distance traveled by train for attaining this velocity. (a). 3 m/s², 1/15 km (b). 2 m/s², 15 km (c). 1/15 m/s², 3 km (d). 4.5 m/s², 2 km
By BYJU'S Exam Prep
Updated on: September 13th, 2023
Step I – It is given that
The initial velocity of train u = 0
Final velocity of train v = 72 km/h = 20 m/s
Time taken t = 5 minutes = 300 s
Consider a as the acceleration and s as the distance covered by the train
Step II – Formula to be used
The first equation of motion
v = u + at
The second equation of motion
s = ut + ½ at2
Where u = Initial velocity
v = Final velocity
a = Acceleration
t = Time
s = Distance covered
Step III – Determine the acceleration
v = u + at
Substitute the values
20 = 0 + a x 300
300a = 20
a = 20/300
a = 1/15 ms⁻2
Step IV – Determine the distance traveled
s = ut + 1/2 at2
Substitute the values
s = 0t + 1/2 (1/15) (300)2
s = 3000 m
s = 3 km
Therefore, the acceleration and the distance traveled by train for attaining this velocity is 1/15 ms⁻2 and 3 km.
Summary:
A train starting from rest attains a velocity of 72 km/h in 5 minutes. Assuming that the acceleration is uniform, find the acceleration and the distance traveled by train for attaining this velocity.
- 3 m/s2, 1/15 km
- 2 m/s2, 15 km
- 1/15 m/s2, 3 km
- 4.5 m/s2, 2 km
A train starting from rest attains a velocity of 72 km/h in 5 minutes. Assuming that the acceleration is uniform, the acceleration and the distance traveled by train for attaining this velocity is 1/15 ms⁻2 and 3 km.
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