Question

A particle is moving with a velocity of 10m/s towards
east. After 10 s its velocity changes to 10m/s
towards north. Its average accelaration is :-
(1) zero
(2) √2m/s² towards N-W
(3) 1/√2m/s² towards N-E
(4) 1/√2m/s² towards N-W​

Answer 1

Hii buddy!

Explanation:

Initial velocity(v1) = 10 i m/s, and

Final velocity(v2) = 10 j m/s.

$average \: acceleration \: = \frac{v2 - v1}{t2 - t1}$

Here, t1 = 0, for initial velocity.

Therefore,

$v2 - v1 = \sqrt{100 + 100}$

$v2 - v1 = 10 \sqrt{2}$

$av.acceleration \: = \frac{10 \sqrt{2} }{10} = \sqrt{2} \: ms^{ - 2}$towards N-W.

SO CORRECT ANSWER IS (2).