A cr travelling at ` 36 km h^(-1)` due North turns West in ` 5 secpmds ` and maintaind the same speed, What is the acceleration of the car ?

Explanation

Refer Fig. 2. (c ). 86. Initial velocity, ` vec v_A`=36 km h^(-1)` due North=10 ms^(-1)` due North.
Find velocity`, vec _B = 36 km h^(-1)= 10 ms^(-1)` due
West. Therefore, ` -vec v_A =- 10 ms^(-1) ` due North
` 10 ms^(-1)` due South.
:. Acceleration = (change in velocity)/(tiem taken)
` vec a= (vec v_B-vec A)/t = (vec v_B+ (- vec V_A)/t`
.
` tan beta = v_B/v_A = (10)/(10) =1 = tan 45^@`
beta = 45^2 ` i.e. South-West direction.