I am told that a torpedo is fired and has an initial velocity of 60 km/hr. After 1 km travelled, its velocity falls to 30 km/hr. We know that the drag force acting on the body is proportional to the speed of the body (at low speeds). Disregarding everything else, I need to find the total distance travelled.
I've tried this:
$$ \frac{d^2y}{dt^2}=k\frac{dy}{dt} \\ \frac{dv}{dt}=kv \\ ln(v)=kt+c_{1} \\ v = 60\ e^{kt}\ , \ (k < 0) \\ \frac{dy}{dt} = 60\ e^{kt} \\ y = \frac{60}{k}(1-e^{kt}) + c_{2} \\ y(0) = 0 \\ y = \frac{60}{k}(1-e^{kt}) $$
So after taking the limit, the distance travelled is $\frac{60}{k}$. Still, I need to find the $k$ to get the exact result. Since this is a Math problem rather than a Physics one, I know I should use the given distance and speed to find $k$, but I can't find where to use it.
Any help is appreciated. Thanks.
We have two expressions, one for speed and one for distance: $$ v=60e^{-kt}\\ y=\frac{60}k\left(1-e^{-kt}\right) $$ Note that we can expand and perform a substitution: $$ y=\frac 1k\left(60-60e^{-kt}\right)\\ y=\frac 1k(60-v) $$ We've already substituted in the initial condition, so let's substitute the other ($v_{y=1}=30$): $$ 1=\frac 1k(60-30)\\ k=30 $$ So the final answer is simply: $$ y_{t=\infty}=\frac{60}k=\frac{60}{30}=2 $$