# Deceleration due to air resistance proportionate to velocity

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?

• To solve $\dot{v}=-kv^2$ with $1$-dimensional $v$, restate that as $\frac{dt}{dv}=-\frac{1}{kv^2}$.
– J.G.
Feb 16, 2022 at 17:03

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.

The initial equation should be:

ma = -𝑘𝑣2