Question

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

• A.
• B.
• C.
• D.

Answer 1

The correct option is C

Let car starts from point A from rest and moves up to point B with acceleration f

Car moves distance BC with this constant velocity in time t $[AS{v}^2=u^2+2as]$

$x=\sqrt{2fS}t.....\left(i\right)[Ass=ut]$

So the velocity of car at point C also will be $\sqrt{2fs}$ and finally car stops after covering distance Y.

Distance $CD\Rightarrow Y=\frac{(\sqrt{2fS}{)}^2}{2\left(f/2\right)}=\frac{2fS}{f}=2S...\left(ii\right)[As{v}^2=u^2-2as\Rightarrow s=u^2/2a]$

so, The total distance AD=AB+BC+CD=15 S(given)

$\Rightarrow$ S+x+2S=15S$\Rightarrow$ x=12S

Substituting the value of x in equation (i) we get


$x=\sqrt{2fS}t\Rightarrow 12S=\sqrt{2fS}t\Rightarrow 144S^2=2fS.t^2$

$\Rightarrow s=\frac{1}{72}f{t}^2.$