Question

to the driver of a car A moving with velocity va=(3i^-4j^)m/s a second car B appears to have a velocity (5i^+12j^)m/s . The true velocity of tje car B is.​

Answer 1

Given:

To the driver of a car A moving with velocity va=(3i^-4j^)m/s a second car B appears to have a velocity (5i^+12j^)m/s.

To find:

True Velocity of car B ?

Calculation:

The velocity vector of car B with respect to car A can be mathematically represented as the vector difference between Velocity vector of car B and car A:

$\rm\therefore \: \vec{v}_{BA} = \vec{v}_{B} - \vec{v}_{A}$

$\rm\implies \: 5 \hat{i} + 12 \hat{j} = \vec{v}_{B} - (3 \hat{i} - 4 \hat{j})$

$\rm\implies \: 5 \hat{i} + 12 \hat{j} = \vec{v}_{B} -3 \hat{i} + 4 \hat{j}$

$\rm\implies \: \vec{v}_{B} = 5 \hat{i} + 12 \hat{j} + 3 \hat{i} - 4 \hat{j}$

$\rm\implies \: \vec{v}_{B} =( 5 + 3) \hat{i} + (12 - 4)\hat{j}$

$\rm\implies \: \vec{v}_{B} =8\hat{i} + 8\hat{j}$

$\rm\implies \: \vec{v}_{B} =8(\hat{i} + \hat{j})$

So, required true Velocity of car B :

$\boxed{ \bold{ \: \vec{v}_{B} =8(\hat{i} + \hat{j})}}$