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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 $28.92\ m$.
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