A boat goes upstream for $3$ hr $30$ min and then goes downstream for $2$ hr $30$ min. If the speed of the current and the speed of the boat in still water are $\frac{10}{3}$ kmph and $\frac{15}{2}$ kmph respectively, how far from its original position is the boat now?

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Speed of boat in still water$=\frac{15}{2}$

Speed of stream $=\frac{10}{3}$

Upstream speed $=\frac{15}{2}-\frac{10}{3}$

Downstream speed $=\frac{15}{2}+\frac{10}{3}$

→ Downstream distance - Upstream distance = far from original.

$=\left(2+\frac{1}{2}\right)\left(\frac{15}{2}+\frac{10}{3}\right) - \left(3+\frac{1}{2}\right)\left(\frac{15}{2}-\frac{10}{3}\right)$

$=\left(\frac{5}{2}\right)\left(\frac{65}{6}\right) - \left(\frac{7}{2}\right)\left(\frac{25}{6}\right)$


$12.5$ km downstream.

Is my approach accurate to find original position of boat?

  • $\begingroup$ Please remove the gap between ":" and "/" to view the image. $\endgroup$ – Scott Kooper Nov 19 '19 at 5:48
  • $\begingroup$ I got the same answer, $12.5~\text{km}$. However, I think it would be wiser to set up a coordinate system. It runs so that the downstream distance is positive. In this case, the upstream speed is actually $\frac{10}{3} - \frac{15}{2}$. The answer remains the same, but some of the signs are flipped in this approach. The expression that I have is $$ \left( \frac{10}{3}- \frac{15}{2} \right) \frac{7}{2} + \left( \frac{10}{3}+\frac{15}{2} \right) \frac{5}{2} = 12.5 $$ $\endgroup$ – Matti P. Nov 19 '19 at 5:57

Your method looks great. As an alternative:

With reference to ground, the water travels downstream for $2.5+3.5=6$ hours.
With reference to water, the boat went upstream for $1$ hour. So the boat is $$ \frac{10}{3} \times 6- \frac{15}{2}\times 1 = 12.5$$ $km$ downstream from the start.

  • $\begingroup$ How boat went upstream for 1 hour only if question clearly says boat goes upstream for 3 hr 30 min? $\endgroup$ – Scott Kooper Nov 19 '19 at 6:47
  • $\begingroup$ What happens if you go north for $3.5$ hours, and then south for $2.5$ hours ? $\endgroup$ – AgentS Nov 19 '19 at 6:48
  • $\begingroup$ Explain me why subtracting $\frac{15}{2}\times 1$ from total distance i.e $\frac{10}{3} \times 6$? $\endgroup$ – Scott Kooper Nov 20 '19 at 9:14
  • $\begingroup$ Alright! Imagine you fell into the water and drowning along the stream. Where would you be in $6$ hours? $\endgroup$ – AgentS Nov 20 '19 at 9:21
  • $\begingroup$ I would be at 6 x whatever stream speed is. $\endgroup$ – Scott Kooper Nov 20 '19 at 10:22

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