Physics Problem

1. May 21, 2005

Silverbackman

Sorry, I acidently posted the same topic in the "General Math" board. Well, it actually is not homework problem necessarily but oh well;

"Someone challenged me a pyhsics problem. The problem is that I don't have time to solve it right now, let alone read it carefully so can someone solve it for me work included? Thanks!

Here is the problem;

""" A car weighing 1000 kg is subject to a force of 100t N for (t is the time in sec) for 3s. What is the speed of the car at the end of 3s, the distance moved, and the work done by the force in the motion. Prove it!!!"""

2. May 21, 2005

Staff: Mentor

:rofl: Good one! You have time to post it, but not to solve it?

Come on, the problem is straightforward. Give it a shot.

3. May 21, 2005

Silverbackman

I did give it a shot, but for some reason my mind is not working as good right now, maybe because I am concentrating on a subject that has nothing to do with physics right now. So in other words, right now I don't get it, I might deep down in my head by right now my mind can't compute it.

Can you do the problem for me so I can see and remember how to do it? Thanks!

4. May 21, 2005

Staff: Mentor

Why don't you show what you've done so far? Start by applying Newton's 2nd law.

5. May 21, 2005

Silverbackman

Okay, I used the forumlas

F=ma ,

a= v/t ,

and

v= d/t

I plugged the info into the euqtions and got

distance=.9M.

speed= .3 M. per second

Are those right?

Also, I still forgot how to find the "work done by the force in the motion" part of the problem.

6. May 21, 2005

Staff: Mentor

No, that's not correct. Note that the force is not constant, thus the acceleration is not uniform. Use Newton's 2nd law to find the acceleration, then integrate to find the speed and distance.

$$a = \frac{dv}{dt} = \frac{d^2x}{dt^2}$$

7. May 21, 2005

Silverbackman

ok so A=Fnet/m is the acceleration of the formula for Newton's 2nd Law. I plugged in like so A=100/1,000= 0.1. Then I don't remeber how you were supossed to use the accelation to find the speed and distance.

8. May 21, 2005

whozum

You have a time varying force which you need to integrate twice to get a position function. You can't do this problem without knowing calculus.

9. May 21, 2005

Silverbackman

I think though if I can look at the work for this problem I might be able to remember. I learned this problem a long time ago, and forgot many things.

If you don't want to post the work for this problem, what about showing me the work of a problem identical to this problem? Or atleast can you post the steps of the problem maybe.

10. May 21, 2005

whozum

Distance travelled: $$\frac{1}{10} \int_0^3 \int_0^t t \ dt\ dt$$

Final velocity: $$\frac{1}{10} \int_0^3 t \ dt$$

11. May 21, 2005

Silverbackman

So did I get the acceleration right?

12. May 21, 2005

whozum

The acceleration is 0.1t, not 0.1
It isnt constant.

13. May 21, 2005

Silverbackman

Acceleration in this case would be the "work done by the force in the motion", right?

Also, what is the speed and distance travelled and how do I use the accleration to find it?

14. May 21, 2005

whozum

I already told you how to do the second part, and the first part makes no sense.
Like I said, if you dont know calculus, you can't do this problem.

15. May 21, 2005

Silverbackman

So then what would the final answers be then? Just post that and I'll see whether I can work backward then.

16. May 21, 2005

whozum

b = acceleration = 0.1
$$a(t) = b t + b_0$$

$$v(t) = \frac{b}{2} t^2 + b_0t + v_0$$

$$x(t) = \frac{b}{6} t^3 + \frac{1}{2} b_0 t^2 + v_0t + x_0|_0^3$$

$$v_0 = x_0 = b_0 = 0$$

$$x(t) = \frac{1}{60} 3^3 - 0 = 9/20 m$$

$$v(t) = \frac{1}{20} 9 = 9/20 m/s$$

$$\int F dx = \int F v dt$$