First slide

Projection Under uniform Acceleration

Question

A 2-m wide truck is moving with a uniform speed v₀ = 8 ms^-1 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

Difficult

A

2.62 ms^-1
B

4.6 ms^-1
C

3.57 ms^-1
D

1.414 ms^-1

Solution

Let the man start crossing the road at an angle 0 with the roadside. For safe crossing, the condition is that the man must cross the road by the time the truck describes the distance (4 + 2 cot $θ$ )
$\frac{4 + 2 cotθ}{8} = \frac{2 / sinθ}{v} or v = \frac{8}{2 sinθ + cosθ} For minimum v, \frac{dv}{dθ} = 0 \frac{- 8 (2 cosθ - sinθ)}{(2 sinθ + cosθ)^{2}} = 0 or 2 cosθ - sinθ = 0 tanθ = 2, so sinθ = \frac{2}{\sqrt{5}}, cosθ = \frac{1}{\sqrt{5}} v_{\min} = \frac{8}{2 (\frac{2}{\sqrt{5}}) + \frac{1}{\sqrt{5}}} = \frac{8}{\sqrt{5}} = 3.57 {ms}^{- 1}$

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