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.

Options

A. `(u +v)/(2)`
B. `(u ^(2) + v ^(2))/( 2)`
C. `sqrt ((u ^(2) + v ^(2))/(2))`
D. `sqrt ((v +u)/(2))`

Answer

C

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 midpoint of a train, while passing the signal is v.. This means that the speed charnges from u to v. in travelling a distance `1/2.`
`therefore v.^(2) -u^(2) =2a ((1)/(2)) =al...(ii)`
Taking ratio of (i) and (ii), `(v ^(2) - u ^(2))/( u .^(2) - u ^(2))=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))`