The initial velocity is u and acceleration (f) is uniform. Final velocity and distance covered in the interval t are v and s respectively. Show that the velocity of the particle at half-distance is more than the velocity for half -time.

Explanation

v = u +ft or, t `= (v-u)/(f)`
Again `v^(2) = u^(2)+2fs " ""or", s = (v^(2)-u^(2))/(2f)`
Velocity at half time,
`v_(1)=u+f""(t)/(2)=u+f""(v-u)/(2f)=u+(v+u)/(2)=(u+v)/(2)`
If the velocity at half-distance is `v_(2)`
`v_(2)^(2)=u^(2)+2f""(s)/(2)=u^(2)+2f"(v^(2)-u^(2))/(4f)`
`=u^(2)+(u^(2)+v^(2))/(2)=(u^(2)+v^(2))/(2)`
`:.v^(2)-v_(1)^(2)=(u^(2)+v^(2))/(2)-(u+v)^(2)/(4)`
`=(1)/(4){2(u^(2)+v^(2))-(u+v)^(2)}`
`(1)/(4) {u^(2)+v^(2)-2uv}`
`=((v-u)/(2))^(2)` , since it is a whole square it is a positive quantity.
So, `v_(1)^(2)-v_(1)^(2) gt 0`
or `v_(2) gt v_(1)`