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

## Homework Statement

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?

## Homework Equations

C1 + C2t + C3t2.

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

## 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.

## The Attempt at a Solution

rock.freak667
Homework Helper
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

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

rock.freak667
Homework Helper
Well for your problem, you just need to know that

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

What does "n" stand for in the equation

rock.freak667
Homework Helper
What does "n" stand for in the equation

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

Wait never mind. Just saw it.

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

rock.freak667
Homework Helper
Using the equation I'm getting -1.24? Is that anything?

That is for the velocity or acceleration?

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

rock.freak667
Homework Helper
I'm actually not sure. Can you use that equation to find both the velocity and acceleration?
And is this the "Simple Power Rule?"

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').

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

Velocity = 22.98 m/s?

Acceleration = -1.24 m/s2?

rock.freak667
Homework Helper
Yes, those should are correct.

Thank you for your help. Very much appreciate it.