# Particle position function

1. Jul 17, 2011

### 1irishman

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

When will the velocity be zero?
t is in minutes
s is in metres

2. Relevant equations

s = 4t^3 - 8t^2 + t - 50

3. The attempt at a solution

I came up with the velocity equation by taking the derivative of function s.
I came up with the acceleration equation by taking the derivative of the velocity function. I don't know how to figure out(solve) when the velocity of the particle will be zero? Help please?
The velocity at t minutes is given by:
v(t) = 12t^2 - 16t + 1
Acceleration is given by:
a(t) = 24t - 16

Last edited: Jul 17, 2011
2. Jul 17, 2011

### Staff: Mentor

Looks good. So plug in the value for v and solve the equation for t.

3. Jul 17, 2011

### 1irishman

can i use quadratic formula, like this?

t = -b + - sqrt b^2 - 4ac/2a

where a = 12, b = -16, and c = 1 ?

4. Jul 17, 2011

### thepatient

Exactly, that will give you the times at which v(t) = 0

5. Jul 18, 2011

### Staff: Mentor

Yes!

6. Jul 18, 2011

### 1irishman

t = 16 + - sqrt208/24

but the answer in the back of book is 4 + - sqrt13/6

I am not sure how they got that?

7. Jul 18, 2011

### Ecthelion

Try simplifying your square root expression more that you already have, and you'll find you have the same answer.

8. Jul 18, 2011

### 1irishman

oh yes thank you i see 16*13 = 208