Finding acceleration through given equation

[dlp211]

Homework Statement

The position of the front bumper of a test car under microprocessor control is given by:
x(t) = 2.17 + (4.80m/s^2)t^2 - (0.100 m/s^6)t^6

Find the acceleration at the first instant when the car has zero velocity.

Homework Equations

The Attempt at a Solution

0m/s^2, 4.8m/s^2

I know that the first instant that the car has zero velocity is at 2.17m at 0 seconds in time. I also know that the car is about to accelerate at this instant. I can't figure out how to determine what the acceleration is though.

 

[Doc Al]

I know that the first instant that the car has zero velocity is at 2.17m at 0 seconds in time.

How did you determine this?

 

[dlp211]

delta x/delta t as t -> 0 is 2.17 (edit: this is wrong). It's because 2.17 is the starting point.

I don't know how to solve 2nd derivatives, I barely know anything about derivatives, our physics course is moving faster then our calc course right now.

 

[Doc Al]

In order to find the velocity as a function of time, you take the first derivative of the position. What does that give you?

Once you have the velocity as a function of time, how would you find the acceleration?

 

[dlp211]

I thought that the derivative function would be 9.6t - .6t^5. But I haven't actually learned derivatives yet.

 

[Doc Al]

I thought that the derivative function would be 9.6t - .6t^5.

Good! That's v(t). And you can solve for the first value of t to make that zero.

Now find a(t).

But I haven't actually learned derivatives yet.

Seems like you know enough to get through this problem.

 

[dlp211]

so a(t) = 9.6 - 3.0t^4?

so a(t) = 9.6 m/s^2?

 

[Doc Al]

so a(t) = 9.6 - 3.0t^4?

Right!

so a(t) = 9.6 m/s^2?

Yes, that's the acceleration at t = 0. Call it a(t = 0), not a(t).

 

[dlp211]

Thanks for the help! My lectures go so fast and I have a hard time following all the formulas.