Question

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

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

Answer 1

Answer: (D)

80 m

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

$v=0,a=\text{constant}$

$⟹s\propto v^2$
So, $\frac{s_2}{s_1}=(\frac{v_2}{v_1}{)}^2\Rightarrow s_2=s_1(\frac{120}{60}{)}^2$$=20\times 4=80m$