Micah rows his boat on a river 4.48 miles downstream, with the current, in 0.32 hours. He rows back upstream the same distance, against the current, in 0.56 hours. Assuming his rowing speed and the speed of the current are constant, what is the speed of the current? 3 miles per hour 8 miles per hour 11 miles per hour 14 miles per hour
@Michele_Laino
@Blonde_Gangsta
@Michele_Laino Hey do u know how to do this ?
8 or 11mph
8 or 11mph
If I call with v_0 the speed of Micah and with V the speed of the current we can write the subsequent equations: \[\Large \left\{ \begin{gathered} \frac{L}{{{v_0} + V}} = 0.32 \hfill \\ \hfill \\ \frac{L}{{{v_0} - V}} = 0.56 \hfill \\ \end{gathered} \right.\]
where L= 4.48
How do u solve that ?
here v_0 is the speed of mica with respect to the water
it is very simple, since those equation compose a system with 2 unknown quantities, so it can be solved with respect to v_0 and V
equations*
can u show me how to solve this plz ?
or do i with me ?
ok!
I can rewrite those equations as below: \[\Large \left\{ \begin{gathered} \frac{{4.48}}{{{v_0} + V}} = 0.32 \hfill \\ \hfill \\ \frac{{4.48}}{{{v_0} - V}} = 0.56 \hfill \\ \end{gathered} \right.\]
do i just multiply
so, we have: \[\Large \left\{ \begin{gathered} \frac{{4.48}}{{0.32}} = {v_0} + V \hfill \\ \hfill \\ \frac{{4.48}}{{0.56}} = {v_0} - V \hfill \\ \end{gathered} \right.\]
so: \[\left\{ \begin{gathered} 14 = {v_0} + V \hfill \\ \hfill \\ 8 = {v_0} - V \hfill \\ \end{gathered} \right.\]
the first one is 14 the second one is 8
so its either B or D
now, I subtract the second equation, from the first one, so I get: \[\begin{gathered} 14 - 8 = \left( {{v_0} + V} \right) - \left( {{v_0} - V} \right) \hfill \\ 6 = 2V \hfill \\ \end{gathered} \]
what is V?
3
that's right! That's your answer!
OMG your awesome .. Thank you very very much :)
thank you! :)
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