Q: A car is moving eastwards with velocity 10 m/s. In 20 sec, the velocity changes to 10 m/s northwards. The average acceleration in this time:

(A)1/√2 m s^{-2} towards N-W

(B)1/√2 ms^{-2} towards N-E

(C) 1/2 m s^{-2 }towards N-W

(D) 1/2 ms^{-2} towards N

Solution: The change in velocity is

$ \displaystyle \vec{\Delta v} = \vec{v_2}-\vec{v_1}$

$ \displaystyle = 10\hat{j}-10\hat{i}$

$ \displaystyle \Delta v = 10\sqrt{2} $ and directed due N – W

Average acceleration = Δv ⁄Δt = 10√2/20

= 1/√2 ms^{-2 }and directed due N-W

Ans: (A)