A car at rest accelerates uniformly at the rate of $4 m / s^{2}$ through a distance of $100 m .$ Find its final velocity and average velocity.

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Solution

$u = 0, a = 4 m s^{- 1}, s = 100 m$
$v^{2} - u^{2} = 2 a s$
$v^{2} = 2 \times 4 \times 100$
$v = \sqrt{800}$
$v = 20 \sqrt{2} m s^{- 1}$
$average velocity = \frac{total distance}{total time}$
$v = u + a t$
$20 \sqrt{2} = 0 + 4 \times t$
$t = 5 \sqrt{2} sec$
So, $v_{a v g} = \frac{100}{5 \sqrt{2}} = 10 \sqrt{2} m s^{- 1}$

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A car at rest accelerates uniformly at the rate of 4m/s2 through a distance of 100m. Find its final velocity and average velocity.

u=0,a=4ms−1,s=100mv2−u2=2asv2=2×4×100v=√800v=20√2ms−1average velocity = total distancetotal timev=u+at20√2=0+4×tt=5√2secSo, vavg=1005√2=10√2ms−1

A car at rest accelerates uniformly at the rate of $4 m / s^{2}$ through a distance of $100 m .$ Find its final velocity and average velocity.