1.

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»

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


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