# Simple dynamic accell problem

1. Sep 6, 2007

### B-80

The acceleration of a particle along an x axis is a = 5.0t, with t in seconds and a in meters per second squared. At t = 2.0 s, its velocity is +13 m/s. What is its velocity at t = 6.5 s?

I dont really know how to go about this, as it is a dynamic acceleration, any help would be greatly appreciated

so far all I have got is a=deltaV/deltaT

2. Sep 6, 2007

### Staff: Mentor

That's OK for average acceleration, but for instantaneous acceleration use a derivative: $a = dv/dt$.

Given acceleration, how do you find velocity? Hint: Use a little calculus.

3. Sep 6, 2007

### B-80

well I would multiply by the time passage, but again this only works for constant acell right? my problem is the derivative, I know how to do them, but i don't understand how that applies. the derivative of a is 5 right? thats great, but what do I do after that

4. Sep 6, 2007

### Staff: Mentor

If you were given v, you'd take its derivative to find a. But you have the opposite situation: You are given a and need to find v. What do you do?

5. Sep 6, 2007

### B-80

integrate? so 2.5t^2=v

6. Sep 6, 2007

### Staff: Mentor

Good! But don't forget the constant of integration.

7. Sep 6, 2007

### B-80

wow awesome thanks a lot.

8. Jan 12, 2008

### Bingo1915

Finding position

I have run across the same type of problem, but I am being asked what the position is. Below is what was given.

Acceleration of a point is a=20t m/s^2. When t=0, s=40 meters and v=-10 m/s. What is the position at t=3 sec?

I've used a=dv/dt to get the velocity, v=80 m/s.

I'm not sure were to start to get the position. Can you advise?

Thank you.

9. Jan 13, 2008

### Staff: Mentor

Using a=dv/dt, you integrated to get the velocity. But v = dx/dt, so integrate once again to get the position.

10. Jan 13, 2008

### Bingo1915

For velocity I got v=10t^2 + c, and pluged 3 in for t and -10 in for c. This gave me v=80 m/s.

If i integrate once more to get position I get s=(10t^3)/3+ct.

Am I correct with the constant?
Plug 3 in for t, and -10 in for c again.

I think Im getting confused with the constant.

11. Jan 13, 2008

### Staff: Mentor

That's the velocity at t = 3; in general it's: v=10t^2 - 10.

But you forgot the new constant of integration. (Hint: You are given the position at t = 0.)

12. Jan 13, 2008

### Bingo1915

Thats what I was forgetting.

Therefore s=(10t^3)/3+ct+d.

Where
t=3
c=-10
d=40

Thanks for the help.