Question

A motor car A is travelling with a velocity of 20 m/s in the north-west direction and another motor car B is travelling with a velocity of
15 m/s in the north-east direction. The magnitude of relative velocity of B with respect to A is
(A)
25 m/s
(B) 15 m/s
(C) 20 m/s
(D) 35 m/s​

Answer 1

Given that,

Velocity of car A = 20 m/s

Velocity of car B = 15 m/s

We need to calculate the magnitude of relative velocity of B with respect to A

Using formula of relative velocity

$|V_{BA}|=|\vec{V_{B}}-\vec{V_{A}}|$

According to vector sum

$\vec{V_{A}}=-\vec{V_{A}}$

So, $\vec{V_{BA}}=\vec{V_{B}}+\vec{V_{A}}$

The magnitude of the relative velocity of B with respect to A is

$V_{BA}=\sqrt{V_{B}^2+V_{A}^2}$

Put the value into the formula

$V_{BA}=\sqrt{15^2+20^2}$

$V_{BA}=25\ m/s$

Hence, The relative velocity of B with respect to A is 25 m/s.

(A) is correct option.

Answer 2

Given:

A motor car A is travelling with a velocity of 20 m/s in the north-west direction and another motor car B is travelling with a velocity of

15 m/s in the north-east direction.  

To find:

The magnitude of relative velocity of B with respect to A is

Solution:

A motor car A is travelling with a velocity of 20 m/s in the north-west direction.

a motor car B is travelling with a velocity of 15 m/s in the north-east direction.  

The relative velocity of B with respect to A is,

|V_{B/A}| = V_{B}bar - V_{A} bar

=  √[V_{B}² + V_{A}²]

= √[15² + 20²]

= √[225 + 400]

= √[625]

= 25.

Therefore, option (A) 25 m/s is the magnitude of the relative velocity of B with respect to A.