# Distance travelled by a car considering only air friction?

Suekdccia
Moved from the technical forums to the schoolwork forums.
TL;DR Summary: Distance traveled by a car considering only air friction?

How much distance would a 3-ton car travel if its initial speed was 17 km/h and we only take into account air's friction? (Assume that the car has an airfoil-like shape, so that the resistance against the air is very low)

I tried to calculate this with the formula Vf² = Vi² + 2·a·d (taking as 0.05 the coefficient of friction of the airfoil-like car against the air) and the resulting distance is 22,84 meters, but it seems too low to me. Am I messing up with something?

Mentor
I tried to calculate this with the formula Vf² = Vi² + 2·a·d (taking as 0.05 the coefficient of friction of the airfoil-like car against the air)
Air resistance is not like friction, it depends on the velocity. What research have you done into solving problems that involve air resistance? Wikipedia might be a good place to start...

Update with a link -- https://en.wikipedia.org/wiki/Drag_(physics)

topsquark and Lnewqban
Homework Helper
Gold Member
You should not use that formula because acceleration is not constant in your case.
Its value will be very high when velocity of the car is still high, but that value will progresively tend to zero at a square ratio as the velocity is degraded by air drag.

topsquark and berkeman
Homework Helper
I tried to calculate this with the formula Vf² = Vi² + 2·a·d (taking as 0.05 the coefficient of friction of the airfoil-like car against the air)
Solving that formula for d gives me about 17 meters.

and the resulting distance is 22,84 meters, but it seems too low to me. Am I messing up with something?
Our numbers do not agree. Yes, in addition to using the wrong formula, you seem to be messing something up. Please show your work.

topsquark
Gold Member
Here is how you arrive at the equation you used:

The work done ##W## is defined as:
$$W = \int Fdx$$
Knowing that ##F=ma##, the work done based on acceleration is:
$$W= \int madx$$
If we want to know the work based on velocity alone:
$$W = \int madx = \int m\frac{dv}{dt}dx = \int m\frac{dx}{dt}dv = \int mvdv$$
Both equations should give the same amount of work, so:
$$\int_{v_i}^{v_f} mvdv = \int_{x_i}^{x_f} madx$$
$$\frac{1}{2}m(v_f^2 - v_i^2) = ma(x_f - x_i)$$
$$v_f^2 = v_i^2 + 2a(x_f - x_i)$$
Which is the equation you used. This assumes that ##F = ma##, where ##a## is constant.

But that is not the case here. The problem identifies the force ##F## that you need to use with the work ##Fdx##. And the work done based on velocity is still ##mvdv##. All you need to do is to equate both as done previously and resolve the integrals. [Hint: the air friction force varies with velocity.]

berkeman and topsquark