# Question #df20b

Dec 28, 2016

Here initial vlocity of the truck is $u = 159.2 \text{km/hr} = \frac{159.2 \times {10}^{3}}{3600} \frac{m}{s}$

Final velocity of the truck $v = 0$

The truck retards uniformly during skidding to come to a stop.

The time of skidding $t = 5.85 s$
So distance of skiding

$s = \frac{1}{2} \left(u + v\right) \times t$

$\implies s = \frac{1}{2} \left(\frac{159.2 \times {10}^{3}}{3600} + 0\right) \times 5.85$

$\implies s = 129.35 m$