Question

A car starting from rest accelerates at rate $\alpha$ through a distance $x$ then continues at constant speed for time $t$ and then decelerates at a rate $\frac{\alpha }{2}$ to come to rest. If the total distance travelled is $15x$. Then

• A. $x=\frac{1}{4}\alpha t^2$
• B. $x=\frac{1}{2}\alpha t^2$
• C. $x=\frac{1}{8}\alpha t^2$
• D. $x=\frac{1}{72}\alpha t^2$

Answer 1

The correct option is D $x=\frac{1}{72}\alpha t^2$

$\left(1\right)v^2-u^2=2\alpha xv=\sqrt{2\alpha x}⟶1x_2=vt=vt⟶2=\sqrt{2\alpha x}t$

and

$0^2-v^2=-2\frac{\alpha }{2}{x}_{3}v=\sqrt{S_2\alpha }⟶3$

now

$x_1+x_2+x_3=15x$

From equation $\left(1\right)$ , $\left(2\right)$ and $\left(3\right)$

$X=\frac{\alpha t^2}{72}$