- #1

spoc21

- 87

- 0

## Homework Statement

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?

## Homework Equations

## 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