Question

An automobile travelling with speed of $60{kmh}^{-1}$ can brake to stop within a distance of $20m$. If the car is going twice, i.e., $120{kmh}^{-1}$, the stopping distance will be

Answer 1

Third equation of motion :

$v^2=u^2+2as$

$0^2=(60\times (\frac{5}{18}){)}^2+2a\times 20$

$0=\frac{2500}{9}+40a$

$a=-\frac{2500}{40\times 9}$

$a=-\frac{125}{18}$

Now, when the initial speed is $120km{h}^{-1}$, acceleration remains same.

$v^2=u^2+2as$

$0^2=(120\times (\frac{5}{18}){)}^2+2s\times \frac{-125}{18}$

$0=\frac{10000}{9}-s\frac{125}{9}$

$s=80$

Therefore, the stopping distance will be 80 m