# 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

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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.

1. 3 m/s2, 1/15 km
2. 2 m/s2, 15 km
3. 1/15 m/s2, 3 km
4. 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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