Question

A car accelerates from rest with $2m/s^2$ on a straight line path and then comes to rest after applying brakes. Total distance travelled by the car is $100m$ in $20$ seconds. Then, the maximum velocity attained by the car is

• A. 10 m / s
• B. 20 m / s
• C. 15 m / s
• D. 5 m / s

Answer 1

Answer: (A)

Given,
Acceleration of car $(\alpha )=2m/s^2$
Distance covered by car in $20$ seconds $(s)=100m$
We know that.
$s=\frac{1}{2}\cdot \frac{\alpha \beta }{\alpha +\beta }{t}^2$
$100=\frac{1}{2}\times \frac{2\beta }{(2+\beta )}\times 20\times 20$
$\frac{100}{400}=\frac{\beta }{(2+\beta )}$
$\Rightarrow \frac{1}{4}=\frac{\beta }{2+\beta }$
$\Rightarrow 4\beta =2+\beta$
$\Rightarrow 3\beta =2$
$\Rightarrow$ retardation $\beta =\frac{2}{3}m/s^2$
$\therefore$ Maximum velocity $\left(v_{\max }\right)=\frac{\alpha \beta t}{\alpha +\beta }$
$\frac{2\times \frac{2}{3}\times 20}{2+\frac{2}{3}}=\frac{\frac{80}{3}}{\frac{8}{3}}=\frac{80}{3}\times \frac{3}{8}$
$=10m/s$