# Drag and time to slow down -__-

1. Oct 26, 2004

### FancyNut

drag and time to slow down... and apparent weight. -__-

A 1.58 m wide, 1.44 m high, 1300 kg car hits a very slick patch of ice while going 20.0 m/s. Air resistance is not negligible.

If friction is neglected, how long will it take until the car's speed drops to 18.0 m/s?

Ok drag is equal to = 1/4 * A * v^2

v^2 = 400 and A = 1.44*1.58

I divided this drag by the mass to get acceleration in the x direction and then used this equation:

$$v_f = v_i + a t$$

t here is delta t...

final velocity is what I want = 18 and initial is 20 while acceleration is the one I got from analyzing the drag force....

I get 11.4 seconds and it's wrong.

I did it again and again to make sure there are no errors but nothing changed.

[EDIT]

btw does anybody know how to get the mass from a chart of apparent weight vs time? I know it's w/g but I only have apparent weight and not weight... I can't tell when it's equal to the real weight. (when a = 0 but I can't tell from the chart).

Here's the chart:

Last edited: Oct 26, 2004
2. Oct 26, 2004

### Tide

The rate of deceleration is not constant as you assumed. Also, where did the factor of 1/4 come from?

3. Oct 26, 2004

### FancyNut

(1/4 * A * v^2 ) <--- equation in my text for drag...

and if deacceleration is not constant... how can I calculate the time?

4. Oct 26, 2004

### Tide

You'll have to integrate

$$\frac {dv}{dt} = - \frac {A}{4M} v^2$$

5. Oct 26, 2004

### FancyNut

ok so integrating that would result in the same drag equation but $$v^2$$ becomes $$\frac {v^3}{3}$$ but since acceleration is not constant I can't use the kinematic equations to solve for delta t...

now that I think about it I'm supposed to get an equation for velocity in t right? after getting that equation I'll just plug the velocity 18 to get t...

[edit2]

ok integrating the derivative of velocity (which is acceleration) gives me the original equation for velocity right? I just plugged in 18 in that equation I integrated and the time is still wrong... :(

Last edited: Oct 26, 2004
6. Oct 26, 2004

### Leong

you integration is wrong; try to rearrange the equation tide gave you so that v terms is together with dv and dt is at the other side of the equation.

7. Oct 26, 2004

### FancyNut

I got several equations and all of them were wrong..

re-arranging that equation gives me this.. right?

$$v = \frac {-4 m}{A t}$$

do I need to integrate that?

integrate the the t?

this sucks. Our course is supposed to be calculus-based but out of 10 HW problems only 1 has to do with calculus... and of course the text barely goes over problems that need calculus. And yes I'm trying to save (a little) face. :(

Last edited: Oct 26, 2004
8. Oct 26, 2004

### Leong

$$\int_{20}^{v} -\frac{4m}{Av^2} dv = \int_{0}^{t} dt$$
Know how to integrate this ?

9. Oct 26, 2004

### FancyNut

no. :(

Is there any other way to solve the problem? the course I'm taking uses VERY easy and simple calculus in the rare occasions it's needed.... so there must be another way. =\

10. Oct 26, 2004

### Leong

there is other way of course but it is more complicated . something like alternative to integration, numerical methods. ok, i will integrate for you . tide must not be very happy with this.
$$\frac{4m}{A}[\frac{1}{v}-\frac{1}{20}]=t$$

11. Oct 26, 2004

### FancyNut

the time I get is still wrong...

bah forget about it and thanks a lot man. I'll ask my teacher next week. The guy hates me but I hope not enough to refuse.

12. Oct 26, 2004

### Leong

is it 12.7 s?

13. Oct 26, 2004

### FancyNut

I got 13.7....

*types in 12.7*

it's correct.

*dies*

 my mistake was rounding/significant figures. and yes I'm not dead. >_<

Last edited: Oct 26, 2004