two.particles a and b are moving with velocities 3m/s and 4m/s in mutually perpendicular direction calculate the magnitude of velocity of A w.r.t B
Answers
Given:
Two particles A and B are moving with velocity 3m/s and 4m/s in mutually perpendicular direction.
To find:
Velocity of A w.r.t B ?
Calculation:
The Velocity of Particle A with respect to Particle B can be mathematically represented as the vector difference between velocity vector of A and B.
\therefore \: \vec{v}_{AB} = \vec{v}_{A} - \vec{v}_{B}∴
v
AB
=
v
A
−
v
B
\implies \: | \vec{v}_{AB} | = \sqrt{ {| \vec{v}_{A} |}^{2} + { | \vec{v}_{B}|}^{2} + 2 |v_{A}| |v_{B}| \cos( {180}^{ \circ} - {90}^{ \circ} ) }⟹∣
v
AB
∣=
∣
v
A
∣
2
+∣
v
B
∣
2
+2∣v
A
∣∣v
B
∣cos(180
∘
−90
∘
)
\implies \: | \vec{v}_{AB} | = \sqrt{ {| \vec{v}_{A} |}^{2} + { | \vec{v}_{B}|}^{2} + 2 |v_{A}| |v_{B}| \cos( {90}^{ \circ} ) }⟹∣
v
AB
∣=
∣
v
A
∣
2
+∣
v
B
∣
2
+2∣v
A
∣∣v
B
∣cos(90
∘
)
\implies \: | \vec{v}_{AB} | = \sqrt{ {| \vec{v}_{A} |}^{2} + { | \vec{v}_{B}|}^{2} + 2 |v_{A}| |v_{B}| (0) }⟹∣
v
AB
∣=
∣
v.
A
∣
2
+∣
v
B
∣
2
+2∣v
A
∣∣v
B
∣(0)
\implies \: | \vec{v}_{AB} | = \sqrt{ {| \vec{v}_{A} |}^{2} + { | \vec{v}_{B}|}^{2} }⟹∣
v
AB
∣=
∣
v
A
∣
2
+∣
v
B
∣
2
\implies \: | \vec{v}_{AB} | = \sqrt{ {(3)}^{2} + {(4)}^{2} }⟹∣
v
AB
∣=
(3)
2
+(4)
2
\implies \: | \vec{v}_{AB} | = \sqrt{ 9 + 16}⟹∣
v
AB
∣=
9+16
\implies \: | \vec{v}_{AB} | = \sqrt{25}⟹∣
v
AB
∣=
25
\implies \: | \vec{v}_{AB} | = 5 \: m {s}^{ - 1}⟹∣
v
AB
∣=5ms
−1
So, velocity of particle A w.r.t particle B is 5 m/s.