Effects of drag on the distance of a car travelling

So:x=\frac{m u^2}{2 F_d}+CAt t=0, x=0, so:x=\frac{m u^2}{2 F_d}At t=10 s, u=45.45 m/s, so:x=\frac{2000*45.45^2}{2*F_d}=657 mIn summary, to calculate the distance traveled by a car with initial speed of 100 m/s and deploying a parachute after 10 seconds, the equation -Fd = m du/dt can be integrated to get the displacement formula x=\frac{m u^2}{2 F_d}. With the given values of mass and initial speed, the resulting
  • #1
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Homework Statement


A car starts at 100 m/s and deploys a parachute. After 10 seconds its speed decreases to 45.45 m/s. Calculate the distance the car has traveled in this duration. The effects of ground resistance are ignored.

drag coefficient and planform area product (Cd * A) = 4 m^2


Homework Equations



drag force, Fd = 1/2 (rho) * u^2 * Cd * A

rho = density, 1.22 kg/m^3
u = speed
Cd = drag coefficient
A = planform area

-Fd = m du/dt

m = mass



The Attempt at a Solution



Basically I managed to calculate the speed after 10 seconds which is 45.45 m/s by integrating -Fd = m du/dt.

I cannot use the SUVAT equations to calculate the distance because it is not constant acceleration (I think). Besides, even if I try this method I get the question wrong. The answer from my lecturer is supposedly s=657m.

I think the above equation must be integrated again to get the displacement but I have tried and I think I'm missing something because I can't manipulate it so that I end up with s in the formula.

Please help as I have pulled my hair off trying to figure this out to no avail.
 
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  • #2
What is the value for mass?
 
  • #3
Oops sorry forgot to put that value. Mass is 2000 kg.
 
  • #4
From:

[tex]ma=F_d[/tex]

And:

[tex]a dx=u du[/tex]

You get:

[tex]\int dx=\int\frac{m u}{F_d}du[/tex]
 

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