A train is moving due East and a car is moving due North, both with the same speed `30 km h^(-1)`. What is the observed speed and diredction of motion of car to the passsenger in the train ?

Explanation

Refe to Fig. 2. (c ). 55,
.
velocity of train, ` vec _T = vec (OA)`,
where ` v_T =30 km h^(-1)`
velocity fo car , `vec v_C= vec (OB)`,
wher ` v_C =30 km h^(-1)`
The obdeved speed and derection of motion of car to the passenger in the rrain will be the relative velocity of car with respect to train. Here
` -vec v_T = vec (OA_1)`. Complete the parallelogtram ` OA_1 CB`.
Now the resultant of ` vec c _C and -vec v_T` will give us the relatice velocity car w.r.t. train `(vec v_(CT))` repreaented by ` vec (OC)`.
Here, ` v_(CT) = sqrt (v_C^2 + v_T^2 -2 v_C v_T cos 90^@)`
`= sqrt (v_C^2 + v_T^2) = sqrt (30^2 = 30^2 0 = 30 sqrt 2 km h^(-1)`
If ` beta` is the angle which ` vec _CT)` makes with the direction of `vec _C` then
` tan beta = (BC)/(OB) = (30)/(30) = 1= tan 45^@ or 45^@ or beta = 45^@`
It means then car is moving West-North direction as observed by passenger in the train.