A train starting from rest attains a velocity of 72 km/h in 5min.Assuming that the acceleration is uniform ,find (1) the acceleration and (2) the distance travelled by the train for attaining this velocity.

vf=at solve for a. put km/h into m/s, and 5min into seconds.

2. d = 0.5a*t^2,

t = 5min. = 300 s.

To find the answer to this question, you need to use the formula for acceleration. The formula for acceleration is:

acceleration = change in velocity / time

Given that the initial velocity of the train is 0 (starting from rest) and the final velocity is 72 km/h, and the time taken to reach this final velocity is 5 minutes.

(1) To find the acceleration:
we need to convert the velocity from km/h to m/s. This can be done by multiplying the velocity by 1000/3600 (since 1 km/h = 1000 m/3600 s).

Final velocity = 72 km/h = (72 * 1000) m/3600 s ≈ 20 m/s
Initial velocity = 0 m/s (starting from rest)

acceleration = (final velocity - initial velocity) / time
= (20 m/s - 0 m/s) / (5 minutes * 60 seconds/minute)
= 20 m/s / 300 s
= 1/15 m/s^2

Therefore, the acceleration of the train is approximately 1/15 m/s^2.

(2) To find the distance travelled by the train:
we can use the formula for distance traveled when the acceleration is uniform:

distance = (initial velocity * time) + (1/2 * acceleration * time^2)

Given that the initial velocity is 0 m/s and the time is 5 minutes (converted to seconds by multiplying by 60):

distance = (0 m/s * 300 s) + (1/2 * (1/15 m/s^2) * (300 s)^2)
= 0 m + (1/2 * 1/15 m/s^2 * 90000 s^2)
= 0 m + (1/30 * 90000 m)
= 3000 m

Therefore, the distance traveled by the train for attaining a velocity of 72 km/h is 3000 meters.