Rain is falling with a velocity `(-4hat(i)+8hat(j)-10hat(k))`. A person is moving with a velocity of `(6hat(i)+8hat(j))` on the ground.
Find the velocity of rain with respect to man and the direction from which the rain appears to be coming.

Options

A. `-6hat(i)-8hat(j)`
B. `4hat(i)-8hat(j)+10hat(k)`
C. `-10hat(i)-10hat(k)`
D. `10hat(i)+10hat(k)`

Answer

D

Explanation

`vec(V)_(rm)=vec(V)_(r)-vec(V)_(m)=(-10hat(i)-10hat(k))`
Ans. `(-10hat(i)-10hat(k))` rain appears to come `45^(@)` with `hat (j)`
`vec(V)_(mr)=-vec(V)_(rm)=10hat(i)+10hat(k)`
Maximum power `=F_(max)xxV`
Maximum force applied by camel is during the accelerated motion.

We have `V^(2)-U^(2)=2as`
`25=0^(2)+2.a.50`
`a=0.25m//s^(2),` for accelerated motion
`:. F_(C)-f=ma`
`:. F_(C)=mu mg +ma=0.1xx1000xx10+1000xx2.5`
`=1000+250=1250N`
This is the critical point just before the point where it attains the maximum velocity of almost `5m//s`
Hence maximum power at this point is `=1250xx5=6250J//s`.