A body with an initial velocity u and a uniform acceleration, covers a distance s in time t and acquires a velocity v. Compare the velocity of the body at half at half of the distance covered with the velocity at half of the total time of travel.

Explanation

Let the acceleration of the body be a.
v = u+at `" ""or", a = (v-u)/(t)`
`:.` Velocity at half of the total time of travel,
`v_(1) = u+a (t)/(2) = u+(t)/(2)((v-u)/(t))`
`= u+(v-u)/(2) = (v+u)/(2)`
Also, `v^(2) = u^(2) +2as " ""or", a = (v^(2)-u^(2))/(2s)`
`:.` Velocity `v_(s)` at half of the distance covered,
`v_(s)^(2) = u^(2) +2a.(s)/(2)=(s(v^(2)-u^(2)))/(2s)=(u^(2)+v^(2))/(2)`
Now, `v_(t)^(2)-v_(s)^(2)=((v+u)/(2))^(2)-((v^(2)+u^(2))/(2))`
`=(v^(2)+u^(2)+2uv)/(4)-(v^(2)+u^(2))/(2)`
`(v^(2)+u^(2)+2uv-2v^(2)-2u^(2))/(4)`
`=((v^(2)+u^(2)-2uv))/(4)=-((u-v)/(2))^(2)`
which is a negative quantity. Hence `v_(1) lt v_(s)`, i.e., the velocity of the body at half-distance is greater than the velocity at half-time.