A stone is dropped from a stationary balloon. It leaves the balloon with zero speed, and \(t\) seconds later its speed \(v\) metres per second satisfies the differential equation \(\frac{\mathrm dv}{\mathrm dt}=10-0.1v^2\).
Find \(t\) in terms of \(v\). Hence find the exact time the stone takes to reach a speed of 5 metres per second.[5]
Find the speed of the stone after 1 second.[3]
What happens to the speed of the stone for large values of \(t\)?[2]
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Answer:(i) \(\frac1{20}\ln|\frac{10+v}{10-v}|+C\). (ii)(a) \(t=\frac12\ln\frac{10+v}{10-v}\), time \(\frac12\ln3\) s; (b) \(7.62\text{ m s}^{-1}\); (c) tends to \(10\text{ m s}^{-1}\).