# Homework Help: Finding the position x(t) with constantly changing acceleration

1. Nov 9, 2008

### alpha01

Im trying to find the position of an object x at time t. There is pretty straight forward formulas to use to find x(t), however acceleration must be fixed... but in this question acceleration is changing at sin2t.

I have been going through period by period working out the new position at x(t-1).. until i finally get to the time i want.

surely there is an easier way to work out the position at time t when acceleration is constantly changing?

2. Nov 9, 2008

### Hootenanny

Staff Emeritus
There is indeed a far more straightforward method to determine x(t), simply consider the definition for acceleration.

$$a = x^{\prime\prime}\left(t\right) = \sin\left(2t\right)$$

3. Nov 9, 2008

### alpha01

thanks, i can see how that is true. However i cant put it all together.

The object is moving at v = -1/2 at time 0, from a starting point of 5 on a grid.. how can i find out where the object is 5 seconds later?

Is it the integral of the integral of acceleration (sin2t) where t = 5?

Last edited: Nov 9, 2008
4. Nov 9, 2008

### Hootenanny

Staff Emeritus
No, try taking it one step at a time. First determine v(t)

$$v\left(t\right) = x^\prime\left(t\right) = \int \sin\left(2t\right) dt$$

5. Nov 9, 2008

### alpha01

(-1/2)cos(2t)

6. Nov 9, 2008

### Hootenanny

Staff Emeritus
Good, but aren't you forgetting something?

7. Nov 9, 2008

### alpha01

woops...

(-1/2)cos(2t) + C

8. Nov 9, 2008

### Hootenanny

Staff Emeritus

9. Nov 9, 2008

### alpha01

C = acceleration at t?

10. Nov 9, 2008

### Hootenanny

Staff Emeritus
No. We have found an equation for the velocity at time t,

$$v\left(t\right) = C - \frac{1}{2}\cos\left(2t\right)$$

Let me rephrase my question: what do the initial conditions tell you about the value of C?

11. Nov 9, 2008

### alpha01

i think that C is the initial velocity of -1/2?

12. Nov 9, 2008

### Hootenanny

Staff Emeritus
Close, but not quite. Try substituting t=0 into the equation for v(t) above.

13. Nov 10, 2008

### alpha01

ok, so using C = -1/2 we get v(0) = -1, and using C = 1/2 we get v(0) = 0

neither of which gets our true initial velocity of -1/2... the only way i found to get the correct initial velocity at t(0) is using C = 0.

14. Nov 10, 2008

### marlon

when t = 0 you get

$$v\left(0 \right) + \frac{1}{2}= C$$

You need to look at your question to find out what the value of the velocity is at t = 0
Once you get that, you'll find a number for C and the velocity is known as a function of time.

Then, you can do the entire same thing :

integrate the formula you found for v(t) to get to x(t)

marlon