A train initially at rest, starts moving and attains a velocity of 72km/hr in 4 minutes. Find :(i) acceleration , (ii) distance covered by the train. {Give answer in m/s terms and dont leave answer in fraction. }
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0
Answer:
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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(i) acceleration
(ii) distance covered by the train.
Initial velocity = 0m/s
Final velocity = 72km/h
Time = 4 sec
where a= acceleration
v= final velocity
u= initial velocity
t= time
s= distance
Firstly change speed km/h to m/s and time min to sec,
Now, for finding the acceleration[a] , use this formula,
For distance covered use second eq' of motion,
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