Question

A car is Going in north direction with velocity 6 meter per second and another car in east direction with velocity 8 meters per second what is relative velocity of A with B also find direction of resultant velocity?.......

Answer 1

• Magnitude of relative velocity is 10m/s
• Direction of relative velocity is 143⁰ from East direction.

EXPLAINATION

Let unit vector along

• East direction is ⃗i
• North direction is ⃗j

GIVEN

• Velocity of first car, ⃗v₁ = 6 ⃗j m/s
• Velocity of second car, ⃗v₂ = 8 ⃗i m/s

CALCULATION

Relative velocity of first car w.r.t. second car

⃗v₁₂ = ⃗v₁ - ⃗v₂

⇒ ⃗v₁₂ = 6 ⃗j - 8 ⃗i

⇒ ⃗v₁₂ = - 8 ⃗⃗i + 6 ⃗j m/s

Magnitude of Relative Velocity

| ⃗v₁₂| = √(8² + 6²)

| ⃗v₁₂| = √(100)

| ⃗v₁₂| = 10 m/s

Direction of Relative Velocity

$tan(θ) = \frac{6}{-8}$

$tan(θ) = -\frac{3}{4}$

$θ = tan⁻¹(-\frac{3}{4})$

$θ = π - tan⁻¹(\frac{3}{4})$

$θ ≈ 180º - 37º$

$θ ≈ 143º$

The direction of relative velocity is 143⁰ from East direction.

Answer 2

Explanation:

Given -

• A car is going in North direction with the velocity of 6 m/s.
• Another car is going in east direction with velocity 8 m/s.

To Find -

• Relative velocity of A with B
• Direction of resultant velocity

Now,

Relative velocity of A wrt B is :-

$v_{1_2} = \vec{v_{1} }- \vec{v_{2}}$

$6 \hat{j} + ( - 8)\hat{i}$

$\implies 6 \hat{j} - 8\hat{i}$

Magnitude of relative velocity :-

v₁₂ = √(-8)² + (6)²

v₁₂ = √64 + 36

v₁₂ = √100

v₁₂ = 10 m/s

Direction of relative velocity :-

⇝tanθ = 6/-8

⇝tanθ = -3/4

⇝θ = tan-¹ (-3/4) from x-axis.