Question

A scooter of mass 120 kg is moving with a uniform velocity of $108\,km\,h^{-1}$. The force required to stop the vehicle in 10 s is

• A. 360 N
• B. 720 N
• C. 180 N
• D. $20 \times 10.8 N$

Answer 1

Answer: (A)

360 N

Answer 2

Force, F = m $\times$ a where a = $\frac{\upsilon - u}{t}$ i.e. F = m$\left(\frac{\upsilon - u}{t} \right)$ Here m = 120 kg, $u = 108 \times \frac{1000}{3600} = 30 \, m \, s^{-1}$, $\upsilon = 0$ and t = 10 s \therefore F = $\frac{120(0-30)}{10} = - 360 \, N$ The force required to stop the vehicle is 360 N. 1 km $h^{-1} = \frac{5 }{18} m \, s^{-1}$