A 2m wide truck is moving with a uniform speed v _ 0 = 8ms^ - 1...

Let the man starts crossing the road at an angle θ 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 θ.


∴ \dfrac{4 + 2 \text\cot \text{\theta}}{8} = \dfrac{2 / \text\sin \text{\theta}}{\text{v}}

\text{or} \text{v} = \dfrac{8}{\text{2} \text\sin \text{\theta} + \text\cos \text{\theta}} \text{...(1)}

$\text{For minimum v,} \dfrac{\text{dv}}{\text{d} \text{\theta}} = 0$

\text{or} \dfrac{- 8 \left(2 \text\cos \text{\theta} - \text\sin \text{\theta}\right)}{( \text{2} \text\sin \text{\theta} + \text\cos \text{\theta} {)}^{2}} = 0

or & & & & & & & & & 2 cos θ - sin θ = 0

or & & & & & & & & & & & & & & & & & tan θ = 2



from Equation (1)

{v}_\min = \dfrac{8}{\text{2} ( \dfrac{2}{5} ) + \dfrac{1}{5}} = \dfrac{8}{5}ms^{- 1}