A particle is moving eastwards with a velocity of 5m/s in 10 seconds the velocity changes to 5m/s northwards. The average acceleration in this time is
(A)21m/s2 towards north-east
(B)21m/s2 towards north.
(C)zero
(D)21m/s2 towards north-west✓ Correct
Answer: (D)
Solution
The velocity turns from 5i^ (east) to 5j^ (north), so the change is Δv=5j^−5i^. The average acceleration is a=tΔv=105j^−5i^=21(j^−i^), of magnitude 2112+12=21 m/s2. Since it points along −i^+j^ (west and north together), the average acceleration is 21 m/s2 directed towards the north-west.
JEE Main · 2005 · 08 May 2005 · Offline · Single correct option