A `2m` wide truck is moving with a uniform speed `v_(0)=8m//s` along a straight horizontal road. A pedestrian starts to cross the road with a uniform speed `v` when the truck is `4m` away from him. The minimum value of `v` so that he can cross the road safely is

Options

A. `1.6 sqrt(5) m//s`
B. `1.2 sqrt(5) m//s`
C. `1.2 sqrt(7) m//s`
D. `1.6 sqrt(7) m//s`

Answer

A

Explanation

Time to cross 2m is `(2/(v sin theta))`….
To avoid an accident
Displacement `=4+v cos thetaxx2/(v sin theta)`
`8xx2/(v sin theta)=4+2 cot theta`
`v sin theta=(16 sin theta)/(4 sin theta +2 cos theta)`
`v_("min")=16/sqrt(4^(2)+2^(2))=1.6 sqrt(5) m//s`
`[ :' (a cos theta + b sin theta) "has max. value "=sqrt(a^(2)+b^(2))]`