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A river boat can travel at $20$ km per hour in still water. The boat travels $30$ km upstream against the current then turns around and travels the same distance back with the current. If the total trip took $7.5$ hours, what is the speed of the current? Solve this question algebraically as well as graphically.

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  • $\begingroup$ Welcome to MSE. What have you tried? Here are some hints: Name some variables. Write down some equations in those variables that amount to distance $=$ rate $\times$ time. $\endgroup$ – Sammy Black Apr 17 '13 at 22:43
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Hint: if the current is $c$, the speed upstream is $20-c$ and downstream is $20+c$

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  • $\begingroup$ :( that's all your goign to give me ahha? $\endgroup$ – user73122 Apr 17 '13 at 22:44
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    $\begingroup$ @user73122: do you have any thoughts on the problem? It is much easier to give a useful answer if we see what you have done. $\endgroup$ – Ross Millikan Apr 17 '13 at 22:45
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$x=(v_{\rm still}-v_{\rm current})t_1$ (when it goes upstream)

$x=(v_{\rm still}+v_{\rm current})t_2$ (when it comes back downstream)

$t_1+t_2=7.5$

and solve for $v_{\rm current}$ , you know the rest of the variables.

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