# Cannot get the value for c?

1. May 4, 2006

### Natasha1

Cannot get the value for c??

The EMF (Effective Motive Force) for a car is the engine driving force minus friction and air resistance.

a)A car of mass 1000Kg travels from rest with constant acceleration to 25 m/s in 30 seconds. Use the impulse-momentum relationship to find the EMF?

b)Now suppose that instead the EMF is made by the driver to increase uniformly during the 30 seconds (by gradually increasing foot pressure on the accelerator). Use the impulse-momentum relationship to find the final speed reached?

c)Compare a) and b)?

Here is the working

The impulse-momentum theorem is:

$$\overline{\sum F} \Delta t = \Delta p$$

where $$\overline{\sum F}$$ is the average net force applied to the object.

a) The acceleration is constant, so the net force applied is constant. I will call the net force F.

$$F \Delta t = \Delta p = m \Delta v$$

$$F = m \frac{ \Delta v}{ \Delta t} = 1000 \frac{25}{30} \, N$$

Thus F = 833 N.

b) The acceleration is now a linear function of time over the first 30 seconds of motion. Thus the net force is also a linear function of time. Using the Calculus version of the impulse-momentum theorem:
$$\int_{t_0}^t F \,dt = \int_{v_0}^v m \, dv$$

The problem I am having here is that the acceleration is a linear function of time, and we don't have the constant: $$a(t) = ct$$ , giving $$F(t)=ma(t)=mct$$

Is there anywhere I have gone wrong?

In terms of c:

$$\int_0^{30}mct \, dt = \int_0^v m \, dv$$

$$\frac{1}{2}mct^2|_0^{30} = mv|_0^v$$

$$450mc=mv$$

$$v = 450c \, \, m/s$$

Last edited: May 4, 2006
2. May 4, 2006

### Natasha1

Damn! I cannot make that [math] writing work :-(

3. May 4, 2006

### neutrino

type "tex" in place of "math".

4. May 4, 2006

### Hootenanny

Staff Emeritus
Replace your math tags with [ tex ] tags, without the spaces. I'm just reading your question now.

~H

5. May 4, 2006

### Jameson

On mathhelpforum.com you use the [math] tag, but here use the [ tex ] tags. :)

6. May 8, 2006

### Natasha1

forgot a vital part of the question

b)Now suppose that instead the EMF is made by the driver to increase uniformly during the 30 seconds (by gradually increasing foot pressure on the accelerator), from 0 to a maximum which is twice the constant value in (a)). Use the impulse-momentum relationship to find the final speed reached?

Last edited: May 8, 2006
7. May 8, 2006

### Hootenanny

Staff Emeritus
You should now be able 2 determine your constant. When t = 30, F = 2 x 833.

~H

8. May 8, 2006

### Natasha1

$$450m*0.0555333333=mv$$

v = 250 ms-1 is this correct?

Last edited: May 8, 2006
9. May 8, 2006

### Hootenanny

Staff Emeritus
Wait a minute!! There is no need for calculus here! What is the impluse - momentum relationship?

~H

10. May 8, 2006

### Natasha1

For par c)Compare a) and b)?

I answered that by gradually increasing foot pressure on the accelerator, the driver has increased by 10 folds the speed of the car. Is there anything else I should add?

11. May 8, 2006

### Hootenanny

Staff Emeritus
I think you've made an error here. Sorry about my previous post, I was look at another topic.

~H

12. May 8, 2006

### Natasha1

I can't see any error, the m cancel out and v is = to what it's = to no?

13. May 8, 2006

### Hootenanny

Staff Emeritus
450 x 0.05553333333 $\approx$ 25

~H

14. May 8, 2006

### Natasha1

I did say that my calculator was playing up

15. May 8, 2006

### Natasha1

For par question c) Compare a) and b)?

I answered that by gradually increasing foot pressure on the accelerator, the driver does not change the speed of the car (over the first 30sec). Is there anything else I should add?

16. May 8, 2006

### Hootenanny

Staff Emeritus
I would imagine that would be fine, unless they ask you to explain the fact you answer is good

~H

17. May 8, 2006

Thanks :-)