Question #df20b

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Answer 1 · 2016-12-28T11:57:24

Here initial vlocity of the truck is $u = 159.2 km/hr = \frac{159.2 \times 10^{3}}{3600} \frac{m}{s}$ $u =159.2 "km/hr"=(159.2xx10^3)/3600m/s$

Final velocity of the truck $v = 0$ $v=0$

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

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

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

$\Rightarrow s = \frac{1}{2} (\frac{159.2 \times 10^{3}}{3600} + 0) \times 5.85$ $=>s=1/2((159.2xx10^3)/3600+0)xx5.85$

$\Rightarrow s = 129.35 m$ $=>s=129.35m$