9. A train starting from rest attains a velocity of 72 km h in 5 minutes. Assuming that the acceleration is uniform, find the (a) acceleration, and (b) distance travelled by the train for attaining this velocity.
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Answer:
acceleration = 1/15 ms-2
distance = 3km
Explanation:
v=u+at
a=(v-u)/t -----------(1)
v=72 km/h
=20m/s
u=0
t=5*60 sec =300sec
by putting values of v,u,t in (1)
a=1/15 m/s2
s=ut +(1/2)at2
=0+(1/2)(1/15)(300)(300)
=3km
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