Question

A river flows due east at 6 ms-1. A boat sails with the
velocity of 8 ms-t due south relative to river. A car goes
on the road along the bank due west at 11 ms-1. Velocity
of car relative to the boat will be
21 ms-1
Ja9 ms-
V353 ms1
1221 ms-1​

Answer 1

For this problem,

Ground is a stationary object.

• Velocity of Boat with respect to ground = $\sf V_{BG}$
• Velocity of River with respect to ground = $\sf V_{RG}$
• Velocity of Car with respect to ground = $\sf V_{CG}$
• Velocity of Boat with respect to river = $\sf V_{BR}$
• Velocity of Car with respect to boat = $\sf V_{CB}$

$\sf \vec{V_{BR}} = \vec{V_{BG}} - \vec{V_{RG}}$

$\longrightarrow \sf \small |\vec{v_{BR}} | = \sqrt{ {|\vec{v_B}|}^{2} + { |\vec{v_R}| }^{2} + 2 |\vec{v_B}| |\vec{v_R}| \cos \theta }$

$\longrightarrow \sf \small |\vec{v_{BR}} | = \sqrt{ {|\vec{v_B}|}^{2} + { |\vec{v_R}| }^{2} + 2 |\vec{v_B}| |\vec{v_R}| \cos90 }$

$\longrightarrow \sf |\vec{v_{BR}} | = \sqrt{ {|\vec{v_B}|}^{2} + { |\vec{v_R}| }^{2} }$

$\longrightarrow \sf | - 8 | = \sqrt{ {|\vec{v_B}|}^{2} + { |6| }^{2} }$

$\longrightarrow \sf 64 - 36 = { |\vec{v_B}| }^{2}$

$\longrightarrow \sf \vec{v_B} = - \sqrt{28} \quad\quad\dots (1)$

Now,

$\sf \vec{v_{CB}} = \vec{v_C} - \vec{v_B}$

$\longrightarrow \sf \small |\vec{v_{CB}} | = \sqrt{ {|\vec{v_C}|}^{2} + { |\vec{v_B}| }^{2} + 2 |\vec{v_C}| |\vec{v_B}| \cos \theta }$

$\longrightarrow \sf \small |\vec{v_{CB}} | = \sqrt{ {|\vec{v_C}|}^{2} + { |\vec{v_B}| }^{2} + 2 |\vec{v_C}| |\vec{v_B}| \cos 90 }$

$\longrightarrow \sf |\vec{v_{CB}} | = \sqrt{ {|\vec{v_C}|}^{2} + { |\vec{v_B}| }^{2} }$

$\longrightarrow \sf \small |\vec{v_{CB}} | = \sqrt{ { | - 11|}^{2} + { | - \sqrt{28} | }^{2} } \quad\quad[From\:(1)]$

$\longrightarrow \sf |\vec{v_{CB}} | = \sqrt{ 121 + 28 }$

$\longrightarrow \sf |\vec{v_{CB}} | = \sqrt{ 149 }$

$\longrightarrow \underline{\underline{\sf{\green{ \vec{v_{CB}} \approx \: 12.21 \:m/s\;in\; south\:-\:west\;direction}}}}$