Question

A train starting from stationary position and moving with uniform acceleration attains a speed of 36 km per hour in 10 minutes. Find its acceleration.

Answer 1

 Initial velocity, u=0m/s
Final velocity, v=36km/h=10m/s
Time, t=10min =10 x 60=600 sec

Average velocity =

`"Inition velocity + Final velocity"/"2"`

Thats is,  Average velocity= `(u+v)/2`

Also,   Distance travelled= Average velocity ×Time

so,     s=`((u+v))/2` × t  ...............(1)

Form the first equation of motion, we have, v =u+at.

put this value of v in equation (1), we get:

        s =`((u+u+at))/2  × t`

or      s=`((2u+at)×t)/2`

or      s = `(2ut+at^2)/2`

or      s = `ut + 1/2 at^2`

where, s =distance travelledby the boby in time t 

          u= initial velocity of the                        body 

and a= acceleration 

Acceleration = `"Final velocity - Initial velocity"/"time taken"`

so,         a= `(v-u)/t=( 10-0)/600 =10/600"m//s^2 =1/60m//s^s = 0.016m//s^2`