Question

A train starting from rest attance a velocity of 72 km hr in 2.5 min assuming that acceraleration is uniform (1) find the acceraleration (2) distance travel by the train.​

Answer 1

Answer:

u=0

v=72 km/hr = 72000/3600= 20m/s

t=2.5 min=150 second

Finding acceleration:-

a= v-u/t

a= 20–0/150

a= 20/150

a= 7.5 m/s²

Finding Distance:-

Using 3rd Equation Of Motion= 2as = v–u²

2as = v²–u²

2×7.5×s= (20)²–(0)²

15×s = 400–0

s = 400/15

s= 26.66 meters