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 `v` so that he can cross the road safely is .

Options

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

Answer

C

Explanation

Time of crosing `=(2)/(vsintheta)`
Time in which truck just able to catch the man`=(4)/(8-vcostheta)`
for safe crosing `(2)/(v sintheta)=(4)/(8-vcostheta)`
or `16-2vcostheta=4vsintheta or v=(4)/(8-v costheta)`
For v minimum `costheta+2sintheta` is maximum
so, `(d)/(d theta)(costheta+2sintheta)=0`
`implies-sintheta+2costheta=0`
`implies tantheta=2`
`impliescostheta=(1)/(sqrt(5)),sintheta=(2)/(sqrt(5))`
`v_(min)=(8sqrt(5))/(5)=(8)/(sqrt(5))`
now `=3.57 m//s`