Question

A car enters a tunnel with velocity 4 m/s ,moves with constant acceleration and comes out of the tunnel with velocity 8 m/s. Velocity of carat the midpointof tunnel is ?
a)5m/s b)6m/s c)3root5m/s d)7m/s​​

Answer 1

Given:

Initial velocity (u) = 4 m/s

Final velocity (v) = 8 m/s

To Find:

Velocity of the car at the mid point of the tunnel

Answer:

In case of constant acceleration mid-point velocity:

$\boxed{ \bf{v_{MP} = \sqrt{ \dfrac{ {u}^{2} + {v}^{2} }{2} } }}$

By substituting values we get:

$\rm \leadsto v_{MP} = \sqrt{ \dfrac{ {4}^{2} + {8}^{2} }{2} } \\ \\ \rm \leadsto v_{MP} = \sqrt{ \dfrac{ 16 + 64 }{2} } \\ \\ \rm \leadsto v_{MP} = \sqrt{ \dfrac{ 80 }{2} } \\ \\ \rm \leadsto v_{MP} = \sqrt{40} \\ \\ \rm \leadsto v_{MP} = 2 \sqrt{10} \: m {s}^{ - 1}$

$\therefore$ Velocity of the car at the mid point of the tunnel = $\sf 2\sqrt{10} \ m/s$