A car travelling with a speed of $15\;m/s$ is braked and it slows down with uniform retardation. It covers a distance of $88 \;m$ as its velocity reduces to $7\;m/s$. If the car continues to slow down with the same rate, after what further distance will it be brought to rest?

Given;

Case I:

Initial Velocity $(u) = 15 \; m/s$

Distance Covered $(s) = 88 \; m$

Final Velocity $v = 7 \; m/s$

Retardation $(a) = \; ?$

We have,

$v^2 = u^2 + 2 \; a \; s$    ⇒   $a = $$\frac{v^2 - u^2}{2\;.\;s} = \frac{7^2 - 15^2}{2 \;* \;88}$ $= -1\;m/s^2$

Case II:

Initial Velocity $(u) = 7 \; m/s $

Retardation $(a) = -1 \; m/s^2$

Final Velocity $(v) = 0 \; m/s$

Distance Covered $(s) = \; ?$

Again we have,

$v^2 = u^2 + 2\;a\;s$

$s = $$\frac{v^2 - u^2}{2 * (-1)}$$ = 24.5\;m$

The car brought to be rest at a distance $24.5 \; m$.

Return to Main Menu