A boat moving in still water slows down so that its speed v (in m/s) after t seconds satisfies \(\displaystyle\frac{dv}{dt} = -kv^2\), where k is a positive constant. Its speed is 20 m/s at t = 0 and 10 m/s at t = 5. Its speed at t = 15 is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 5 m/s
The variables separate:
\( \displaystyle \frac{dv}{v^2} = -k\,dt \)
Integrating both sides:
\( \displaystyle -\frac{1}{v} = -kt - c \)
\( \displaystyle \frac{1}{v} = kt + c \)
At t = 0, v = 20:
\( \displaystyle c = \frac{1}{20} \)
At t = 5, v = 10:
\( \displaystyle \frac{1}{10} = 5k + \frac{1}{20} \)
\( \displaystyle k = \frac{1}{100} \)
At t = 15:
\( \displaystyle \frac{1}{v} = \frac{15}{100} + \frac{1}{20} = \frac{20}{100} = \frac{1}{5} \)
\(v = 5\)
Why the other options are incorrect
Option (1) 2.5 m/s — Incorrect. This assumes the speed halves every 5 seconds; it is \(\frac{1}{v}\) that grows by equal steps, not the speed that halves.
Option (2) \(\displaystyle\frac{20}{3}\) m/s — Incorrect. This is the speed at t = 10, where \(\frac{1}{v} = \frac{15}{100}\).
Option (4) 0 m/s — Incorrect. This assumes the speed falls by 10 m/s every 5 seconds. Since \(\frac{1}{v} = \frac{t}{100} + \frac{1}{20}\) stays finite, v never becomes 0.