# Acceleration, Velocity and Position (calculus)

1. Mar 16, 2010

1. The problem statement, all variables and given/known data

The acceleration of a particle is directly proportional to time, t. At t = 0, the velocity of that particle, v, = 16 in./sec. Knowing that v = 15 in/sec and x = 20in when t = 1, determine the velocity, acceleration and position at x = 7.

2. Relevant equations

3. The attempt at a solution

I took
$$\frac{dv}{dt} = kt, so \: v = \frac{k}{2}t^2 + C$$

Solving for C:
$$16 = \frac{k}{2}(0)^2 + C$$
C = 16

Solve for k

$$1 = \frac{k}{2}(1)^2 + 16$$

k = -30

Therefore:

$$v(t) = \-15*(t)^2 + 16$$

This is already wrong; v(7) = -719 in/sec but the book states the answer is -33 in/sec

Any advice on what's gone wrong would be greatly appreciated; this is a revision question too...

Cheers,

2. Mar 16, 2010

### Gregg

k is -2

3. Mar 16, 2010

### ehild

You need to find the velocity when x = 7. Find x(t) first.

ehild

4. Mar 16, 2010

Ah, thankyou ehild; I failed to read the question once again... I will give it a shot. How is k = -2?

5. Mar 16, 2010

### ehild

Looks OK.

6. Mar 16, 2010

I found x(t) by integrating and finding C. I obtained this:

$$x(t)=16t-5t^3+9$$

Then solving for t when x = 7

$$16t-5t^3+2 = 0$$

Sorry, how does one solve this?

Cheers,

Last edited: Mar 16, 2010
7. Mar 16, 2010

### ehild

Don't forget that k=-2, so v(t)=16-t^2.

Looking at the solution for v(t) at x=7 , it should be rather t=7 instead of x. Otherwise why would they ask the position at x=7?

ehild

Last edited: Mar 16, 2010
8. Mar 16, 2010

Sorry, this is probably really stupid, but where/how does k = -2?

9. Mar 16, 2010

### ehild

"v = 15 in/sec and x = 20in when t = 1"

15=k/2*1^2+16

ehild

10. Mar 16, 2010