By what factor does the drag force on a car increase as its velocity increases from 65.0 kilometers per hour to 110 kilometers per hour?
In this exercise, we want to solve for the ratio of the drag force at the higher speed to the drag force at the lower speed. We’ll call the drag force at the higher speed 𝐹 sub 𝑑𝐻 and the drag force at the lower speed 𝐹 sub 𝑑𝐿. It’s their ratio that we want to solve for.
As we begin, let’s recall a relationship between drag force and velocity. The drag force that acts on an object is proportional to the square of that object’s velocity. In our scenario, we have a higher velocity, we’ll call it 𝑣 sub ℎ, of 110 kilometers per hour and a lower velocity, we’ll call 𝑣 sub 𝑙, of 65.0 kilometers per hour. By the relationship between drag force and velocity, we can see that 𝐹 sub 𝑑𝐻 in our scenario is proportional to 𝑣 sub ℎ squared, and that 𝐹 sub 𝑑𝐿 is proportional to 𝑣 sub 𝑙 squared. If we divide the proportionality for the higher value by the proportionality for the lower value, then we see that the drag force will increase by a factor of the velocity at the higher speed to the velocity at the lower speed squared. In other words, 𝐹 sub 𝑑𝐻 divided by 𝐹 sub 𝑑𝐿 is equal to 𝑣 sub ℎ divided by 𝑣 sub 𝑙 quantity squared.
We can now insert the given values for these two velocities. 110 kilometers per hour divided by 65.0 kilometers per hour quantity squared is equal to 2.86. That’s the factor by which the drag force increases as the velocity increases.