# Question f53ce

Feb 26, 2018

a)$81000 k m {h}^{-} 2 = \textcolor{red}{6.25 m {s}^{-} 2}$ is the deceleration of the car
and
b) $4$ seconds is the time elapsed before he stops.

#### Explanation:

Initial velocity of vehicle is $u$ =$90 k m {h}^{-} 1$
$\textcolor{red}{= 90 \times \frac{1000 m}{3600 s} = 25 m {s}^{-} 1}$

After applying brakes, he can stop in Distance $s = 50 m = \frac{50}{1000} = 0.05 k m$

$\textcolor{red}{= 50 m}$

Final velocity $v = 0 k m {h}^{-} 1$ as the vehicle stops.

(a) To find the deceleration of the car:
Apply equation:

$s = \frac{{v}^{2} - {u}^{2}}{2 a}$

$\implies a = \frac{{v}^{2} - {u}^{2}}{2 s}$

$a = \frac{{0}^{2} - {90}^{2}}{2 \times 0.05}$

$\textcolor{red}{= \frac{0 - {25}^{2}}{2 \times 50}}$

$a = - \frac{8100}{0.1} = - 81000 k m {h}^{-} 2$

$\textcolor{red}{= - \frac{625}{100} = - 6.25 m {s}^{-} 2}$

$\therefore$Deceleration = $81000 k m {h}^{-} 2$

$O R \textcolor{red}{\implies 81000 \times \frac{1000 m}{3600 s \times 3600 s} = 6.25 m {s}^{-} 2 = \text{deceleration}}$

Note : We give negative sign if acceleration is to be written and positive sign if the word deceleration or retardation is to be written.

(2) To find the time elapsed before he stops:

Apply equation :

$v = u + a t$

$\implies t = \frac{v - u}{a}$ $\frac{k m h {r}^{-} 1}{k m h {r}^{-} 2}$

$t = \frac{0 - 90}{-} 81000$

$t = 0.00 \overline{111} h r . = 0. 0 \overline{6}$ minutes =  4

$\sec o n \mathrm{ds}$color(red)(= -25/-6.25 = 4s)#

$\therefore$ a)$81000 k m {h}^{-} 2 = \textcolor{red}{6.25 m {s}^{-} 2}$ is the deceleration $\textcolor{red}{O R}$ $\textcolor{red}{\text{accceleration} = - 6.25 m {s}^{-} 2}$of the car
and
b) $4$ seconds is the time elapsed before he stops.