Question

A,B,C are three objects each moving with constant velocity. A's speed is 10m/s in a direction PQ . The velocity of B relative to A is 6m/s at an angle of cos -1 (15/24) to PQ. The velocity of C relative to B is 12m/s in a direction QP, then find the magnitude of the velocity C.

Answer 1

Velocity of A, Va= 10

Velocity of B, Vba= 6 cosy i +6 siny j

Given is $cosY=\frac { 15 }{ 24 }$ from this we can find $sinY=\frac { \left( \sqrt { 351 } \right) }{ 24 }$

Now Vba =Vb-Va $\Rightarrow 6 cosy i +6 siny j =VB- 10i$

So Vb=10i + 6 cos y i +6siny j

$=\frac { 55 }{ 4i } +\frac { \left( \sqrt { 351 } \right) }{ 4j }$

Now $Vcb=Vc-Vb \Rightarrow -12i=Vc-(\frac { 55 }{ 4i } +(\frac { \left( \sqrt { 351 } \right) }{ 4j } )$

$Vc=\frac { 7 }{ 4i } +\frac { \left( \sqrt { 351 } \right) }{ 4j }$

So now the magnitude of $Vc=\sqrt { { \left( \frac { 7 }{ 4 } \right) }^{ 2 }+{ \left( \frac { \left[ \sqrt { 351 } \right] }{ 4 } \right) }^{ 2 } } =\left( \frac { 7 }{ 4 } \right) +\left( \frac { 18.7 }{ 4 } \right) =\frac { 25.7 }{ 4 } =\frac { 26 }{ 4 }$

= 6 m/s