A motorboat moves along a straight line path along a river bank from point A to point B. If the speed of the boat in still water is doubled, then the trip from A to B and back again would take 20% of the time that the motorboat usually takes for the journey. By how many times is the actual speed of the motorboat higher than the speed of the river flow?

Question

Prepp Team · Accepted Answer

Motorboat Speed Ratio Calculation

This problem involves calculating the ratio between a motorboat's speed in still water and the river's flow speed, based on changes in travel time.

Let $ v_b $ be the speed of the motorboat in still water, and $ v_s $ be the speed of the river flow. Let $ d $ be the distance between points A and B.

Original Journey Time Calculation

The total time $ T $ for a round trip (A to B and back to A) depends on speeds downstream and upstream.

Speed downstream (with the current): $ v_b + v_s $
Speed upstream (against the current): $ v_b - v_s $
Time downstream: $ t_{down} = \frac{d}{v_b + v_s} $
Time upstream: $ t_{up} = \frac{d}{v_b - v_s} $
Total original time: $ T = t_{down} + t_{up} = \frac{d}{v_b + v_s} + \frac{d}{v_b - v_s} $
Combine terms: $ T = d \frac{(v_b - v_s) + (v_b + v_s)}{(v_b + v_s)(v_b - v_s)} = d \frac{2v_b}{v_b^2 - v_s^2} $

Modified Journey Time Calculation

The motorboat's speed in still water is doubled to $ 2v_b $.

New speed downstream: $ 2v_b + v_s $
New speed upstream: $ 2v_b - v_s $
New time downstream: $ t_{down, new} = \frac{d}{2v_b + v_s} $
New time upstream: $ t_{up, new} = \frac{d}{2v_b - v_s} $
Total new time: $ T_{new} = t_{down, new} + t_{up, new} = \frac{d}{2v_b + v_s} + \frac{d}{2v_b - v_s} $
Combine terms: $ T_{new} = d \frac{(2v_b - v_s) + (2v_b + v_s)}{(2v_b + v_s)(2v_b - v_s)} = d \frac{4v_b}{(2v_b)^2 - v_s^2} = d \frac{4v_b}{4v_b^2 - v_s^2} $

Relating the Times

The problem states the new round trip time $ T_{new} $ is 20% of the original time $ T $.

$ T_{new} = 0.20 \times T = \frac{1}{5} T $

Substitute the expressions for $ T $ and $ T_{new} $:

$ d \frac{4v_b}{4v_b^2 - v_s^2} = \frac{1}{5} \left( d \frac{2v_b}{v_b^2 - v_s^2} \right) $

Cancel $ d $ and $ v_b $ (since $ v_b \neq 0 $):

$ \frac{4}{4v_b^2 - v_s^2} = \frac{2}{5(v_b^2 - v_s^2)} $

Solving for the Speed Ratio

Cross-multiply to solve for the relationship between $ v_b $ and $ v_s $:

$ 4 \times 5(v_b^2 - v_s^2) = 2 \times (4v_b^2 - v_s^2) $ $ 20(v_b^2 - v_s^2) = 8v_b^2 - 2v_s^2 $ $ 20v_b^2 - 20v_s^2 = 8v_b^2 - 2v_s^2 $

Rearrange the equation:

$ 20v_b^2 - 8v_b^2 = 20v_s^2 - 2v_s^2 $ $ 12v_b^2 = 18v_s^2 $

Find the ratio $ \frac{v_b^2}{v_s^2} $:

$ \frac{v_b^2}{v_s^2} = \frac{18}{12} = \frac{3}{2} $

Take the square root to find the ratio $ \frac{v_b}{v_s} $:

$ \frac{v_b}{v_s} = \sqrt{\frac{3}{2}} $

Conclusion

The actual speed of the motorboat is $ \sqrt{\frac{3}{2}} $ times higher than the speed of the river flow.

Prepp Team · Answer

$\sqrt{\frac{2}{3}}$

Prepp Team · Answer

$\frac{2}{3}$

Prepp Team · Answer

$\frac{3}{2}$