First slide

Relative motion in 2 dimension

Question

A 2-m wide truck is moving with a uniform speed $v_{0} = 8 {ms}^{- 1}$ 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 y 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 $θ$ 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 $θ$ )
$\begin{array}{l} So, \frac{4 + 2 \cot θ}{8} = \frac{2 / \sin θ}{v} or v = \frac{8}{2 \sin θ + \cos θ} \\ For minimum v, \frac{d v}{d θ} = 0 \end{array}$
$\begin{aligned} \begin{array}{l} or \frac{- 8 (2 \cos θ - \sin θ)}{(2 \sin θ + \cos θ)^{2}} = 0 or 2 \cos θ - \sin θ = 0 \\ or \tan θ = 2, so \sin θ = \frac{2}{\sqrt{5}}, \cos θ = \frac{1}{\sqrt{5}} \\ v_{min} = \frac{8}{2 (\frac{2}{\sqrt{5}}) + \frac{1}{\sqrt{5}}} = \frac{8}{\sqrt{5}} = 3.57 {ms}^{- 1} \end{array} \end{aligned}$

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