Question

A particle is moving eastwards with a velocity of 15 m/s. suddenly it move towards north and moves with same speed in a time 10 sec. the average acceleration during this time is
Please answer the following question

Answer 1

Answer:

Explanation:

According to the problem the particle is moving eastwards with the velocity of 15 m/s

Therefore the initial velocity is v1 = 15 m/s

Then it moves towards north with the same velocity

Therefore the final velocity, v2 = 15 m/s

Therefore change in velocity,

Δv = v2 – v1 = 15 m/s due north - 15 m/s due east

or,Δv = 15 m/s due north + 15 m/s due west

Δv = (v2^2 + v1^2)^1/2

Δv = 15√2 m/s

Direction of Δv is given by,

tanθ = v2/v1 = -1

=> θ = -45°

Therefore the acceleration is a = Δv/t =  15√2/10 = 3/√2 m/s2 NW

Answer 2

Answer:

The answer is 3) 3/root 2