A train is moving with a velocity of ` 30 km h^(-1)` due East and a car is moving with a veloicty of ` 40 km h^(-1)` due North . What is the velocity of car as appears to a passenger in the train ?

Explanation

In Fig. 2 (c ) . 89, velocity of train, ` v_t = 30 km h^(-1)` ltbRgt represented by ` vec (OA)`. Velocity of car, ltbRgt ` v_c = 40 km h^(-1)` represented by ` vec (OB)`. We have to find relatve velocity of car w.r.t. train. To bring the train at rest, apply equal and opposite velocity of train on car, which is ` 30 km h^(-1)`
represented by ` vec (OC)`. Then the relative velocity of car w.r.t. train is represented by ` vec (OD) i.e. the diagonl of the reocity of car w.r.t. train is
.
` v_(CT) = OD = (OB^2 + BD^2)^(1//2)`
`= ( 40^2+30^2)^(1//2) = 50 km h^(-1)`
Let ` /_DOB= theta`, then ` tan theta = BD// OB`
` = 30 // 40 = 3//4 =0. 75 = tan 36^@52'` ltbRgt or ` theta 36^@ 52'` West of Noth.