I stumbled with this problem in a notebook that has been bothering me...The answer is written but there's no explanation nor a steb-by-step procedure or anything. If you know how to analyse the problem and create an equation from this, I will be very grateful!
A motor boat is moving at 10km/h being on tranquil waters. While moving, the motor was turned off and the speed of the boat went down to 6km/h.
A)Considering that the resistance force of the water is proportional to the speed of this, find the speed of the boat at 2 minutes of turning the motor off.
B)Find the travelled distance of the boat of 1 minute of having turned the motor off.
Answers: V= 0.467km/hr Distance = 85.2m
As given, the resistant force is:$$ f = -k \cdot \dot {v}$$ By Newton's second law, the equation of motion is:$$\dot{v}+\alpha\cdot {v} =0$$ where, $\alpha={k \over m}$, $m$ is the mass of the object that is not provided. We can have the solution of the differential equation in the format:$$v=C_1 \cdot e^{\lambda t}$$ To satisfy the equation of motion, $\lambda=-\alpha$.
With $v(0)=10$, and $v(t_2)=6$, we have $C_1=10$ and $C_1 \cdot e^{-\alpha t_2}=6$.
Therefore, $$e^{-\alpha t_2}={6 \over 10}$$
We must have the value of $t_2$ to determine the value of $\alpha$, but it is not provided in the original question.
Inversely, according one of the answer, $t_2$ must be 20 seconds.
We now have:$$v=10 \cdot e^{-\alpha t}$$
where $\alpha=180\ln({5\over3})$
Integrate $v$ to $x$, we have:$$x=\int_0^t{v}dt={10 \over \alpha}(1-e^{-\alpha t})$$