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Q.
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
Laws of Motion
Solution:
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 $h^{-1} = \frac{5 }{18} m \, s^{-1}$