A bike riding at $22.4\ m/s$ skids to come to a halt in $2.55\ s$. Conclude the skidding distance of the bike.

Here,

Initial velocity of the bike, $u=22.4\ m/s$

Final velocity of the bike, $v=0\ m/s$

Let the acceleration be $a$ of the bike. 

Time taken, $t=2.55\ sec$

Therefore, $a=\frac{v-u}{t}=\frac{0-22.4}{2.55}$

Or $a=-8.7\ m/s^2$

Now using the equation $s=ut+\frac{1}{2}at^2$

Skidding distance, $s=22.4\times 2.55+\frac{1}{2}( -8.7)\times ( 2.55)^2$

$=57.12-28.2$

$=28.92\ m$

Thus, the skidding distance of the bike is $28.92\ m$.
 

Updated on: 10-Oct-2022

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