Question

the velocity of a particle moving on a circular path is 5cm/s toward north at any instant After traversing one-fourth of the path its velocity is 5cm/s toward east .Indicate the change in velocity in a vector diagram​

Answer 1

Answer:

$5\sqrt{2}  cm/s$

Explanation:

Let $v_1$ be the velocity of particle towards North and $v_2$ be the velocity of the particle towards East

The change in velocity is

Δv=$v_2-v_1$

According to vector subtraction

Δv=$\sqrt{(v_1)^2+(v_2)^2}$=$\sqrt{5^2+5^2}$=$5\sqrt{2}  cm/s$

Answer 2

Answer:

this is the answer

Explanation:

Hope it will help