Question

The Relative Velocity Of Car A With Respect To Carb Is 30 Root 2 M/S Due North East And Velocity Of Car B Is 20M/S Due South . The Relative Velocity Of Car C With Respect To Car A Is 10 Root 2 M/S Due North-West . The Speed Of Car C And Direction (In Term Of Angle It Make With The East)

Answer 1

Answer:

The velocity of C is $20\sqrt{2}$ m/s in North-East direction (45° towards North from East)

Explanation:

If we take East direction as +x axis and North direction as +y axis then

The relative velocity of the car A w.r.t. car B is $30\sqrt{2}$ m/s in North-East direction

i.e.

$V_{A/B}=30\sqrt{2}\cos45^\circ \hat i+30\sqrt{2}\sin 45^\circ\hat j$

$V_{A/B}=30\sqrt{2}\times \frac{1}{\sqrt{2}}\hat i+30\sqrt{2}\times \frac{1}{\sqrt{2}}\hat j$

$\implies V_{A/B}=30\hat i+30\hat j$

Velocity of car B

$V_B=-20\hat j$

Velocity of car C w.r.t. car A

$V_{C/A}=-10\sqrt{2}\cos45^\circ \hat i+10\sqrt{2}\sin 45^\circ\hat j$

$V_{C/A}=-10\sqrt{2}\times \frac{1}{\sqrt{2}}\hat i+10\sqrt{2}\times \frac{1}{\sqrt{2}}\hat j$

$V_{C/A}=-10\hat i+10\hat j$

From the convept of relative velocty

$V_{A/B}=V_A-V_B$

And $V_{C/A}=V_C-V_A$

Adding the above two

$V_{A/B}+V_{C/A}=V_C-V_B$

$\implies V_C=V_{A/B}+V_{C/A}+V_B$

$\implies V_C=30\hat i+30\hat j+(-10\hat i+10\hat j)-20\hat j$

$\implies V_C=20\hat i+20\hat j$

Magnitude of the velocity of C

$=\sqrt{20^2+20^2}$

$=20\sqrt{2}$

Direction of velocity

$\theta=\tan^{-1}\frac{20}{20}$

$\implies \theta=\tan^{-1}1=45^\circ$

Thus the velocity of C is $20\sqrt{2}$ m/s in North-East direction

Hope this helps.

Answer 2

Explanation:

If we take East direction as +x axis and North direction as +y axis then

The relative velocity of the car A w.r.t. car B is 30\sqrt{2}302 m/s in North-East direction

i.e.

V_{A/B}=30\sqrt{2}\cos45^\circ \hat i+30\sqrt{2}\sin 45^\circ\hat jVA/B=302cos45∘i^+302sin45∘j^

V_{A/B}=30\sqrt{2}\times \frac{1}{\sqrt{2}}\hat i+30\sqrt{2}\times \frac{1}{\sqrt{2}}\hat jVA/B=302×21i^+302×21j^

\implies V_{A/B}=30\hat i+30\hat j⟹VA/B=30i^+30j^

Velocity of car B

V_B=-20\hat jVB=−20j^

Velocity of car C w.r.t. car A

V_{C/A}=-10\sqrt{2}\cos45^\circ \hat i+10\sqrt{2}\sin 45^\circ\hat jVC/A=−102cos45∘i^+102sin45∘j^

V_{C/A}=-10\sqrt{2}\times \frac{1}{\sqrt{2}}\hat i+10\sqrt{2}\times \frac{1}{\sqrt{2}}\hat jVC/A=−102×21i^+102×21j^

V_{C/A}=-10\hat i+10\hat jVC/A=−10i^+10j^

From the convept of relative velocty

V_{A/B}=V_A-V_BVA/B=VA−VB

And V_{C/A}=V_C-V_AVC/A=VC−VA

Adding the above two

V_{A/B}+V_{C/A}=V_C-V_BVA/B+VC/A=VC−VB

\implies V_C=V_{A/B}+V_{C/A}+V_B⟹VC=VA/B+VC/A+VB

\implies V_C=30\hat i+30\hat j+(-10\hat i+10\hat j)-20\hat j⟹VC=30i^+30j^+(−10i^+10j^)−20j^

\implies V_C=20\hat i+20\hat j⟹VC=20i^+