# Velocity and acceleration calculus

1. Jul 27, 2008

### brizer

Hi! This is a Calculus I problem following the movement of a particle. I got stuck at part c, but I included the other parts in case I messed up somewhere along the way.

The problem:
An object is moving on the x-axis with displacement given by x(t)=-t3 +12t2 -27t +34

a) Find the equations for the velocity and the acceleration of the particle at any time t.
b) Find the velocity and acceleration at t=1, t=4, t=6, and t=7.
c) At each of the times for part (b), state whether x is increasing or decreasing and at what rate.
d) At each of the times from part (b), state whether the particle is speeding up or slowing down. Justify your answer.
e) At what time(s) in the interval [0,10] is x a relative maximum. Justify your answer.

My attempts:
a) the velocity = x'(t) = -3t2 +24t -27
the acceleration = x''(t) = -6t + 24
b) x'(1) = -6. x''(1) = 18
x'(4) = 21. x''(4) = 0
x'(6) = 9. x''(6) = -12
x'(7) = -6. x''(7) = -18
c) here's where I ran into trouble. I think the velocity tells me whether the function is increasing or decreasing and the acceleration tells me at what rate. However, with that explanation, my answers don't make sense.
at t=1, the function decreases at a rate of 18
at t=4, the function increases at a rate of 0
at t=6, the function increases at a rate of -12
at t=7, the function decreases at a rate of -18
d) I think the particle would be speeding up when the acceleration is positive, so on the interval (0,4).
e) I haven't done this part yet, but I think the maximum can be found using the first and second derivatives. When the 1st derivative = zero there is a maximum or minimum and when the 2nd derivative is negative the point is a maximum.

If you could help me out, I'd really appreciate it!

2. Jul 27, 2008

### konthelion

If x(t) is increasing on the interval I if $$f(x_{1}) < f(x_{2})$$
If x(t) is decreasing on the interval I if$$f(x_{1}) > f(x_{2})$$

So, if you find the instantaneous velocity at t, you can tell if it is increasing or decreasing by the sign. If the v(t) > 0, then it is increasing. If the v(t) < 0, then it is decreasing.

For example, at t = 1, substitute t into your equation v(t)=x'(t) = -3t2 +24t -27.

The rate of change is simply the value for v(t), the instantaneous velocity at t.

Set the derivative of x(t) which is x'(t)=v(t) to 0 to obtain critical values and determine the local/relative max min by using the second partial derivative test (i.e. your acceleration function a(t).

If a(t) > 0, then it is a minimum
If a(t) < 0, then it is a maximum

3. Jul 28, 2008

### brizer

Thanks! That makes sense, but I have one last question on part d.

If the particle speeds up when a(t) is positive and slows when a(t) is negative, what is it doing when a(t)=0? I know that a(4)=0, so should I take the third derivative to find out if it's increasing or decreasing?

4. Jul 28, 2008

### konthelion

When a(t) = 0, it means that the velocity is constant, so it is neither speeding up nor slowing down.