equations of motion

Question

Stopping distance for particle in KN-010 is:

RankGuru · Accepted Answer

From KN-010, $u = 10 	ext{ m/s}$ and $a = -2 	ext{ m/s}^2$.
Using the third equation of motion:
$v^2 = u^2 + 2 a s$
Since the particle stops, $v = 0$:
$0 = 10^2 + 2(-2)s$
$4s = 100 \implies s = 25 	ext{ m}$.
Alternatively, using $s = u t + \frac{1}{2} a t^2$ with $t = 5 	ext{ s}$ found in KN-010:
$s = (10 	imes 5) - \frac{1}{2} 	imes 2 	imes 25 = 50 - 25 = 25 	ext{ m}$.

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