Question

A train starting from rest attains a velocity of 72km/hr in 5 min. Assuming that acceleration is uniform. Find(i) acceleration(ii) the distance travelled by the train for attaining this velocityby the equation 2as=v^(2)-u^(2)

Answer 1

u = o m/s

v = 72 km/h = 72*5/18 m/s

= 20 m/s

t = 5 min = 5*60 s

= 300 sec

a) a = ?

As v = u + a t

so, 20 = 0 + a * 300

so, a = 20/300 = 1/15 m/s^2

= 0.0667 m/s^2

b) s = ?

As 2as=v^(2)-u^(2)

so, 2* 0.0667 * s = 20^2 - 0

so, s = 400/ 0.1333 m

= 3000.075 m