A train is moving with a velocity ` 40 km h^(-1)` due East and a car is moving with a velocity ` 60 km h^(-10` due North, What is the velocity of train as appears to a passenger in the car ?

Explanation

Refer to Fig,2 (c ). 88 ,
.
Velocity of train, ` vec _T = vec (OA),
where ` v_T = 40 km h^(-1)`
Velocity of car, ` vec v_c = vec (OB)` ,
Velocity of train as appears to a passenger in the car is the relative velocity of train w.r.t. car ltbRgt ` is, ` vec v(TC)= vec _T - vec _C= vec _T + (- vec _C)` ltbRgt ` = vec (OA) + vec (OB_1) = vec (OC)`
wher ` | vec v_(TC)| = sqrt v_T^2 +v_C^2) = sqrt (40^2 + 60^2)`
` = 72 .11 km h^(-1)` ltbRgt If ` vec _(TC)` make and angle ` beta` with East direction, ltBrgt then ` tan beta = v_C/v_T = ( 60)/( 40) = 3/2 = tan 55^@ 19'`
` beta = 56^@ 19' ` South of East .