Question

An automobile travelling with a speed of $60km/h$, can apply brake to stop within a distance of $20m$. If the car is going twice as fast i.e., $120km/h$, the stopping distance will be

• A. $60m$
• B. $40m$
• C. $20m$
• D. $80m$

Answer 1

The correct option is D $80m$

Here we use $3^{rd}$ equation of motion,
$v^2=u^2+2as,$where $v=$ final velocity$u=$ initial velocity,
$a=$ acceleration due to gravity$,s=$ Distance or height.

For first case $u_1=0m/s,v_1=60km/h=\frac{5}{18}\times 60m/s=\frac{50}{3}m/s,a_1=a,s_1=d_1=20m$

$0=v_1^2-2\times a\times d_1\Rightarrow v_1^2=2a{d}_1.............\left(1\right)$

For second case $u_2=0m/s,v_1=120km/h=\frac{5}{18}\times 120m/s=\frac{100}{3}m/s,a_2=a,s_2=d_2$

$0=v_2^2-2a{d}_2\Rightarrow v_2^2=2\times a\times d_2\Rightarrow 4v_1^2=2a{d}_2..............\left(2\right)$

When equation $\left(2\right)$ is divided by equation $\left(1\right)$ we get,
$\frac{4v_1^2}{v_1^2}=\frac{2\times a\times d_2}{2\times a\times d_1}\Rightarrow \frac{d_2}{d_1}=4$
$d_2=4\times d_1\Rightarrow d_2=4\times 20\Rightarrow d_2=80m$