# Position of a Particle as Function of Time (Velocity, Acceleration, etc.)

1. Aug 28, 2010

### 09jml90

1. The problem statement, all variables and given/known data
The position of a particle as a function of time (in s) is given by C1 + C2t + C3t2.
Let C1 = 19.0 m, C2 = 39.1 m/s, and C3 = -0.62 m/s2.

(a) What is the velocity of the particle at time t = 13.0 s?
(b) What is the particle's acceleration at time t = 13.0 s?

2. Relevant equations

C1 + C2t + C3t2.

Let C1 = 19.0 m, C2 = 39.1 m/s, and C3 = -0.62 m/s2.

3. The attempt at a solution

I wasn't sure how to do this problem from the beginning, so I tried playing around with some numbers to see if anything could come up. I tried substituting 13.0 into the formula to create a quadratic equation and got: -104.78x2 + 508.3x + 19. I then used the quadratic formula to get some points: (-0.037, 0.0495) (4.888, 0.06186). If I'm doing anything right so far, I would say from the graph that the velocity is decreasing and the acceleration is negative?

Any help would be greatly appreciated. Thank you.
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Aug 28, 2010

### rock.freak667

So you have (after substituting the Cs) s=19+39.1t-0.62t2.

You need to get 'v' and 'a' when 't=13'. All you need to do is remember your Relevant Equations

v=ds/dt and a=dv/dt

3. Aug 28, 2010

### 09jml90

I actually haven't taken calculus and don't know derivatives yet but I'm taking physics and calculus simultaneously right now. I'm going to try to look derivatives up and see what I can come up with and post my solutions.

Thanks for the tip. I'll let you know what I get

4. Aug 28, 2010

### rock.freak667

Well for your problem, you just need to know that

y=tn then dy/dt=ntn-1 for all n.

5. Aug 28, 2010

### 09jml90

What does "n" stand for in the equation

6. Aug 28, 2010

### rock.freak667

'n' is a constant so it can be any number.

7. Aug 28, 2010

### 09jml90

Wait never mind. Just saw it.

8. Aug 28, 2010

### 09jml90

Using the equation I'm getting -1.24? Is that anything?

9. Aug 28, 2010

### rock.freak667

That is for the velocity or acceleration?

10. Aug 28, 2010

### 09jml90

I'm actually not sure. Can you use that equation to find both the velocity and acceleration?
And is this the "Simple Power Rule?"

11. Aug 28, 2010

### rock.freak667

I am not sure if that is the name of the rule.

But yes, you will use the initial equation to find the velocity and acceleration.

Remember v=ds/dt (so you will now have a formula for 'v') and then a =dv/dt (you will get a formula for 'a').

12. Aug 28, 2010

### 09jml90

Oh I didn't even think about substitution. Man I suck at this more than I thought. One sec.

13. Aug 28, 2010

### 09jml90

Velocity = 22.98 m/s?

Acceleration = -1.24 m/s2?

14. Aug 28, 2010

### rock.freak667

Yes, those should are correct.

15. Aug 28, 2010

### 09jml90

Thank you for your help. Very much appreciate it.