A man is walking on a level road at a speed of 3.0 km/h. Rain drops fall vertically wth a sped of 4.0 km/h. Find the velocity of the raindrops with respect to the man.

Explanation

We have to find the velocity of raindrops with respect to the man. The velocity of the rain as well as the velocity of theman are given with respect to the street. We have
`vecv(rain,man)=vecv_(rain,street)-vecv_(man,street)`
ure shows the velocities.

it is clear the ure that
`v_(rain,man)=sqrt((4.0 km/h)^2+(3.0 km/h)^2)`
`=5.0 km/h`
The angle with the vertical is theta, where `tan theta =(3.0 km/h)/(4.0 km/h)=3/4`
Thus, the rain appears to fall at an angle `tan^-1 (3/4)` with the speed 5.0 km/h as viewed by the man.
The relation between teh accelertions measured from two frames can be obtasined by differentiating equation with respect to time.
we have,
`d/dt(vecv_(P,s)=d/dt(vecv_(P,S)+d/dt(vecc(S,S)`
or `veca_(P,S)=veca_(P,S)+veca_(S,S)`
If S' moves with respect to S at a uniform velocity, `veca_(S,S)=0 and so`
`veca-(P,S)=veca_(P,S)`
If two frames are moving with respect to each other with uniform velocity, acceleration of a body is same in both the frames.