A train is moving with constant acceleration. When the ends of a train pass by a signal their speeds are u and v respectively. Calculate the speed of the midpoint of the train while passing the signal.

Explanation

Suppose length of a train is l. Now, speeds of both ends are different, so we can say that the speed changes from u to v in travelling distance l. Hence, using the equation of motion `v ^(2) - v_(0) ^(2) = 2ax,` putting `v=v, v _(0)=u and x = l,` we get `v^(2) -u ^(2) =2al " "...(i)`
Now, suppose the speed of the midoint of a train, while passing the signal is v.. This means that the speed changes from u to v. in travelling a distance `1/2.`
`therefore v .^(2) -u ^(2) =2a ((1)/(2)) = al ...(ii)`
Taking rato of (i) and (ii),` (v ^(2) - u ^(2))/(v^(.2) - u ^(2))`
`therefore 2 (v^(.2) - u ^(2)) = v ^(2) -u ^(2)`
`therefore v^(.2) - u ^(2) = (v ^(2) - u ^(2))/( 2)`
`therefore v ^(.2) = (v ^(2)- u ^(2) )/(2) + u ^(2)`
`therefore v ^(.2) = (v ^(2) + u^(2))/( 2)`
`therefore v.= sqrt ((v ^(2) + u ^(2))/(2))`