# Finding acceleration of a particle

1. Sep 18, 2009

### Amel

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

The location of a particle (in m) is given by its x, y and z coordinates as function of the time (in s) as:

x = -33+29t and y = -11-27t+7t2 and z = 95-9t-5t2

What the magnitude of the object's acceleration at t = -10.00s?

3. The attempt at a solution

Ok so im not sure how to do this. my friend was telling me you take the derivative of the three components but we haven't learned how to do that in calc yet. So im not sure how to do it.

If I remember he was telling me you take the exponents and multiply it by the coeficent so fo x you get 0 + 29 and do that untill all the Ts are gone and then take the square root of it.

Which doesent really make sense and I got the wrong answer when I did that.

How do I solve this problem.

2. Sep 18, 2009

### 206PiruBlood

$$\stackrel{d}{dt}t^{n}=nt^{n-1}$$

This is the general formula to calculate the derivative of your functions.

Keep in mind that a derivative is a rate of change. The rate of change of your position function is velocity, and the velocity rate of change is acceleration.

Last edited: Sep 18, 2009
3. Sep 18, 2009

Yes. Acceleration is the 2nd derivative of the displacement function (location).

Fortunately, yours is a simple one. You will need to know two rules;

1) The derivative of a constant (a number) is zero

Example: $\frac{d}{dt}[10]=0$. Said, "The derivative of '10' with respect to the variable t is zero.

2) Derivatives of functions of the form tn=n*tn-1

Example: $\frac{d}{dt}[3t^6]=6*3*t^{6-1}=18t^5$

So, do this to each of the 3 functions above once and the result will be the particle's velocity. Then do it again to all 3 velocity functions and the result will be the particle's acceleration as a function of t.

4. Sep 18, 2009

### drizzle

then you can get the object's acceleration from:

a=√[(ax)2+(ay)2+(az)2]

ps. I don’t think t is minus, is it?!