Question

12. The speed of a motor boat with respect to water is v = 14 m/s. The speed of water with
respect to the banks is 6 m/s. When the boat began travelling upstream, a buoy was dropped from
it. A buoy is a body that can float in water. The boat travelled 6.3 km upstream (with respect to
banks), turned about and caught up with the buoy. Find the time T lapsed between dropping the
buoy and catching up with it again

Answer 1

Answer on Question #42034 – Physics – Mechanics | Kinematics | Dynamics

1. The speed of a motor boat with respect to water is $\mathrm{v} = 14 \, \mathrm{m/s}$. The speed of water with respect to the banks is $6 \, \mathrm{m/s}$. When the boat began travelling upstream, a buoy was dropped from it. A buoy is a body that can float in water. The boat travelled $6.3 \, \mathrm{km}$ upstream (with respect to banks), turned about and caught up with the buoy. Find the time $T$ lapsed between dropping the buoy and catching up with it again.

Solution.

The speed of the buoy equals to the speed of the river.

When the boat goes upstream, its speed with respect to banks is $v - u$. When the boat goes downstream, its speed with respect to the banks is $v + u$.

One can make an equation of the equality of time, which was elapsed by the buoy and by the boat (the boat goes $s$ upstream and $(s + uT)$ downstream):

$$T = \frac{s}{v - u} + \frac{s + uT}{v + u}.$$

One can find the time $T$ :

$$T = \frac{2s}{v - u}.$$

Let check the dimension: $\left[T\right] = \frac{m}{\frac{m}{s}} = s$.

Let evaluate the quantity: $T = \frac{2 \cdot 6300}{14 - 6} \approx 1575(s) \approx 26.3(\min)$.

Answer: $26.3 \, \mathrm{min}$.

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