Solving a Physics Problem: Car's Speed, Distance, and Work in 3 Seconds

[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!"""

 

[Doc Al]

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?

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

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

 

[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!

 

[Doc Al]

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

 

[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.

 

[Doc Al]

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}

 

[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 acceleration to find the speed and distance.

 

[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.

 

[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.

 

[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

 

[Silverbackman]

So did I get the acceleration right?

 

[whozum]

The acceleration is 0.1t, not 0.1
It isn't constant.

 

[Silverbackman]

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

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

 

[whozum]

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

 

[Silverbackman]

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

 

[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

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

Answers:

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

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

I'm not certain about the work function to give an answer.

 

[Doc Al]

The work done can be found using:
\int F dx = \int F v dt