Question

A train starting from rest attains a velocity of 72 km per hour in 5 minute gyming at the acceleration is uniform find the acceleration and distance travelled by the train for attaining this velocity

Answer 1

u=0
v=72km/hr
In m/s=72*5/18
= 20m/s
a=v-u/t
=20-0/300
a=1/15m/s^2
S=ut+1/2at^2
=0*1/15+1/2*1/15*300*300
=1/2*1/15*90000
=1/30*90000
=3000m
=3km

Answer 2

initial velocity (u) = 0 (as it starts from rest)
final velocity (v) = 72km/h
= 20m/s
time taken (t) = 5min.
= 5 × 60s
= 300s

using 1st equation of motion
acceleration (a) = (v-u)/t
= 20/ 300
= 1/15 m/s^2

using second equation of motion
distance (s)= ut +(1/2)(a)(t)^2
=(0)(1800) +(1/2)(1/15)(300)(300)
= 0 + 3,000
= 3,000

so the distance covered by that train was 3,000m.