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Question: A 2 m wide truck is moving with a uniform speed v<sub>0</sub> = 8 m/s along a straight horizontal ro...

A 2 m wide truck is moving with a uniform speed v0 = 8 m/s along a straight horizontal road. A pedestrian starts to cross the road with a uniform speed v when the truck is 4 m away from him. The minimum value of v so that he can cross the road safely is –

A

2.62 m/s

B

4.6 m/s

C

3.57 m/s

D

1.414 m/s

Answer

3.57 m/s

Explanation

Solution

Let the man starts crossing the road at an angle q as shown in figure. For safe crossing the condition is that the man must cross the road by the time the truck describes the distance 4 + AC or 4 + 2 cot q.

\ 4+2cot⁡θ8\frac{4 + 2\cot\theta}{8}=2/sin⁡θv\frac{2/\sin\theta}{v}

or v =82sin⁡θ+cos⁡θ\frac{8}{2\sin\theta + \cos\theta} … (1)

For minimum v, dvdθ\frac{dv}{d\theta}= 0

or −8(2cos⁡θ−sin⁡θ)(2sin⁡θ+cos⁡θ)2\frac{- 8(2\cos\theta - \sin\theta)}{(2\sin\theta + \cos\theta)^{2}}= 0

or 2 cos q – sin q = 0

or tan q = 2

From Eq. (1) vmin =82(25)+15\frac{8}{2\left( \frac{2}{\sqrt{5}} \right) + \frac{1}{\sqrt{5}}}=85\frac{8}{\sqrt{5}}= 3.57 m/s