# Homework Help: Velocity/acceleration using derivatives (answer check)

1. May 13, 2010

### spoc21

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

An object is traveling along a linear path according to the equation
s(t) = 4t^3 - 3t^2 + 5 where t is measured in seconds and s(t) measured in meters.

1. How fast is the object moving at t = 4 seconds?

2. What is the position of the object when it stops moving?

3. How far has the object traveled when its acceleration is zero?

4. Is the object moving towards or away from the origin at t = 3 seconds?

2. Relevant equations

3. The attempt at a solution

1. How fast is the object moving at t = 4 seconds?

Over here, I substituted the value of t=4 into the original equation, s(t)

s(4) = 4(4)^3 - 3(4)^2 + 5
=213

so 213 m/s? is this correct?

2. What is the position of the object when it stops moving?

For this, I took the first derivative, and let it equal to 0:

s(t) = 4t^3 - 3t^2 + 5
s' (t)=(4)(3) t^(3-1)-(3)(2) t^(2-1)+0
s' (t)=12t^2-6t

12t^2-6t=0
t= 0 or 1/2

so it stops moving after 0.5 seconds? is this correct?

3. How far has the object traveled when its acceleration is zero?

For this, I took the second derivative, and made it equal to 0:

s' (t)=12t^2-6t
s'' (t)=24t-6

24t-6 = 0

t = 0.25 or 1/4

Acceleration is zero at 0.25 seconds

Now I substitute the value of t=0.25 into the original equation s(t):

s(t) = 4t^3 - 3t^2 + 5
s(0.25) = 0.0625- 0.1875 + 5
= 4.875.

So, the object has traveled 4.25 m when its acceleration is zero? is this correct

4. Is the object moving towards or away from the origin at t = 3 seconds?

This was very confusing.

I substituted the value of t = 3 into both the original equation, and the first derivative(velocity):

s(t) = 4t^3 - 3t^2 + 5
s(3) = 108 - 27 + 5
= 86

s' (t)=12t^2-6t
s' (3)=108-18
=90

Because both the values of the speed, and the velocity are positive, the object is moving away from the origin? is this correct?

I would really appreciate it if some one could take a look over my working, and point out any mistakes.

Thank you

2. May 13, 2010

### Staff: Mentor

s(4) = 213 represents where the object is, not how fast it is going. The units are in meters.
Assuming it started moving at t = 0 sec. and stopped moving at t = .5, the question asks for the position of the object when it stops moving.
How did you get 4.25 m?
Since s'(3) = v(3) > 0, the object is moving away from the origin. If you graph the derivative you will see that at some times, the object is moving toward the origin.

3. May 13, 2010

### LCKurtz

No. s(4) is the position in meters of the particle when t = 4. It tells you where it it, not how fast it is moving.

At times t = 0 and 1/2 the object is instantaneously at rest. I wouldn't say it "stops moving". That would imply to me that it stays in that spot thereafter, which it doesn't. Maybe it is just momentarily reversing direction.

That just gives the position at t = 1/4. Where was it when t = 0? Is the distance traveled the difference between s(1/4) and s(0) or maybe did the object reverse direction? You need to check these things.
That part looks correct.

4. May 13, 2010

### spoc21

ok, how could I find the speed..would I need to divide it by 4s?

yes, thats a typo. is the answer 4.875 correct?

But is my assumption correct? should I just put down that the object is moving away from the origin at t= 3?

Thanks!

5. May 13, 2010

### Staff: Mentor

1. You have a formula for the velocity. Use it to find v(4). Divide by 4s???? I don't get what you mean by that at all.
2. Yes, although as LCKurtz pointed out, the object stops only momentarily.
3. Yes. Since the velocity (which is different from speed) is positive at t = 3, the object is moving away from the origin. Don't say just that it's moving away from the origin. It's probably best to give justification for your statement.