Deceleration due to air resistance proportionate to velocity

Question

I'm given a question to suppose that an ice skater on a frictionless surface is attempting to reduce his speed. His initial speed is $v_0$. His deceleration (negative acceleration) is purely due to air resistance, given by $F=-kmv^2$, where $m$ is the skater's mass and $k$ is a constant. From the time when he begins to decelerate, what is his speed at time $t$?

Obviously, firstly I recognise the force of deceleration $F=ma$ is such that $ma=-kmv^2$ Therefore, $$a=-kv^2$$

To find the speed at time $t$, it should be such that: $$\int a \ dt = \int -kv^2 \ dt $$ $$v=-kv^2t$$

But what I'm confused is that the acceleration/deceleration at a given time already depends on its speed which means my previous equation is already wrong since I cannot treat $v$ as a constant. How am I supposed to take this into consideration to find the speed?

Answer 1

Indeed, you cannot solve it that way because $v$ is a function of time. Instead you treat it as a differential equation: $$a = \frac{d v(t)}{dt}=-k v^2(t)$$

This has the solution $$v(t)=\frac{1}{kt+C}$$ and we can solve for $C$ by using the initial condition $v(0)=v_0$ from which we get $$v(t)=\frac{1}{kt+1/v_0}$$

You can plug that back in to check that it satisfies both the differential equation and the initial condition.

Answer 2

The initial equation should be:

ma = -𝑘𝑣2

Then apply answer 1.

Air resistance is a force, not an acceleration. It is not dependent on the mass of the object - just coefficient of drag, density of the air, and cross sectional area of the object. This equation above maintains a mass term in the solution, which makes sense - more massive objects will experience less deceleration at the same air resistance.