Question

A car starts from rest and moves in a straight line with a constant acceleration $\alpha$. After time $t_0$, brakes are applied which causes retardation of magnitude $\beta$ and car finally stops. The distance travelled by the car is

• A. $\alpha t_0^2\frac{(\alpha +\beta )}{4\beta }$
• B. $\alpha t_0^2\frac{(\alpha +\beta )}{2\beta }$
• C. $\beta t_0^2\frac{(\alpha +\beta )}{2\alpha }$
• D. $\beta t_0^2\frac{(\alpha +\beta )}{4\alpha }$

Answer 1

The correct option is B $\alpha t_0^2\frac{(\alpha +\beta )}{2\beta }$

Distance traveled in $t_0$ seconds

$S_1=0\times t_0+\frac{1}{2}\alpha t^2$(u=0 which is at rest)

Final velocity after $t_0$ seconds,

$V_s=0+\alpha t_0$

Final velocity after applying retardation $\beta =0$

Initial velocity =final velocity after $t_0$ seconds

$0=(V_s{)}^2-2\beta (s_2)$ ($s_2$ is the distance traveled after $\beta$ retardation)

$s_2=\frac{(V_s{)}^2}{2\beta }$

$s_2=\frac{{\alpha }^2{t}_0^2}{2\beta }$

$s_1+s_2=\frac{1}{2}\alpha t_0^2(\alpha +\frac{\alpha }{\beta })$

$=\frac{\alpha t_0^2(\alpha +\beta )}{2\beta }$