A `2-m` wide truck is moving wigh 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 `v` so that he can cross the road safely is .

Options

A. `2.62 ms^(-1)`
B. `4.6 ms^(-1)`
C. `3.57 ms^(-1)`
D. `1.414 ms^(-1)`

Answer

C

Explanation

Let the man start crossing the roat at an angle `theta` with the roadside. For safe crossing, the condition is that the man must cross the road by the time the truck describes the destance `4+ 2 cot theta`
.
So, `(4+2 cot theta)/(8) =(2// sin theta)/(v)` or `v =8 (2 sin theta + cos theta)`
For minimum `v,(vd)/(d theta) =0`
or `(-8( 2 cos theta -sin theta)/((2 sin theta + cos theta))^(2) =0` or `2 cos theta -sin theta =0`
or `tan theta =2`, so `sin theta =(2)/(sqrt5), cos theta =(1)/(sqrt5)`
`v_(min) =(8)/(2((2)/(sqrt50) +(1)/(sqrt5)) =(8)/(sqrt5) =3.57 m s^(-1)`.