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Sindrei [870]
1 year ago
8

Shareen finds that when she drives her motorboat upstream she can travelwith a speed of only 8 m/s, while she moves with a speed

of 12 m/s whenshe heads downstream. What is the current of the river on which Shareenis traveling?
Physics
1 answer:
ANEK [815]1 year ago
5 0

Let vb be the velocity of the motorboat and let vs be the velocity of the stream.

We know that when she drives upstream the velocity is 8 m/s, in this scenario the velocities point in opposite directions, then we have the equations:

v_b-v_s=8

When she drives downstream the velocites point in the same direction then we have the equation:

v_b+v_s=12

hence we have the system of equations:

\begin{gathered} v_b-v_s=8 \\ v_b+v_s=12 \end{gathered}

Solving the first equation for the velocity of the boat we have:

v_b=8+v_s

Plugging this in the second equation we have:

\begin{gathered} 8+v_s+v_s=12 \\ 2v_s=4 \\ v_s=\frac{4}{2} \\ v_s=2 \end{gathered}

Therefore, the velocity of the stream is 2 m/s

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960-y is the distance traveled in free fall

v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 16\times y+0^2}\\\Rightarrow v^2=32y\ m/s

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v^2-u^2=2g(960-y)\\\Rightarrow 0-32y=2g(960-y)\\\Rightarrow -32y=2\times -9.81(960-y)\\\Rightarrow 960-y=\dfrac{-32}{2\times -9.81}y\\\Rightarrow 960-y=1.63098878695y\\\Rightarrow 960=2.63098878695y\\\Rightarrow y=\dfrac{960}{2.63098878695}\\\Rightarrow y=364.881828749\ m

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s=ut+\frac{1}{2}at^2\\\Rightarrow 364.881828749=0t+\frac{1}{2}\times 16\times t^2\\\Rightarrow t=\sqrt{\frac{364.881828749\times 2}{16}}\\\Rightarrow t=6.75353452598\ s

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