A car, moving with a velocity of $36 k m / h,$ accelerates uniformly at a rate of $2.5 m / s^{2}$ for $5$ seconds. Find the distance covered during this period.

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Solution

$u = 36 k m / h r = 10 m s^{- 1}$
$a = 2.5 m s^{- 2}, t = 5 sec$
$v = u + a t$
$v = 10 + (2.5 \times 5)$
$= 22.5 m s^{- 1}$
$v^{2} - u^{2} = 2 a s$
$\Rightarrow \frac{{22.5}^{2} - 10^{2}}{2 \times 2.5} = s$
$s = 81.25 m$

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A car, moving with a velocity of 36km/h, accelerates uniformly at a rate of 2.5m/s2 for 5 seconds. Find the distance covered during this period.

u=36km/hr=10ms−1a=2.5ms−2,t=5secv=u+atv=10+(2.5×5) =22.5ms−1v2−u2=2as⇒22.52−1022×2.5=ss=81.25m

A car, moving with a velocity of $36 k m / h,$ accelerates uniformly at a rate of $2.5 m / s^{2}$ for $5$ seconds. Find the distance covered during this period.

$u = 36 k m / h r = 10 m s^{- 1}$
$a = 2.5 m s^{- 2}, t = 5 sec$
$v = u + a t$
$v = 10 + (2.5 \times 5)$
$= 22.5 m s^{- 1}$
$v^{2} - u^{2} = 2 a s$
$\Rightarrow \frac{{22.5}^{2} - 10^{2}}{2 \times 2.5} = s$
$s = 81.25 m$