# acceleration motion problem.

• Dec 3rd 2006, 03:14 PM
acceleration motion problem.
I got most of them this weekend. There's just a few I'm not sure about.

1. NASA launches a rocket at http://hosted.webwork.rochester.edu/...388a5f95d1.pngseconds. Its height, in meters above sea-level, as a function of time is given by http://hosted.webwork.rochester.edu/...34bda8d651.png.
Assuming that the rocket will splash down into the ocean, at what time does splashdown occur?

The rocket splashes down after ______ seconds.

How high above sea-level does the rocket get at its peak?

The rocket peaks at _______ meters above sea-level.

2.Solve the equation http://hosted.webwork.rochester.edu/...b0c587f2f1.png by factoring.
I know that one of the solutions is definitly -8. The other one I'm getting 5/4 but it says that's wrong.

3.Complete the equation of the circle centered at (-1,-1) with radius 6.
I'm getting 34=0 for this but I'm not to sure what the equation for a cirlce is.

4.A vendor sells ice cream from a cart on the boardwalk. He offers vanilla, chocolate, strawberry, blueberry, and pistachio ice cream, served on either a waffle, sugar, or plain cone. How many different single-scoop ice-cream cones can you buy from this vendor?
Do I use nPr or nCr for this one? And which number equals n and which one equals r.

5. The angle of elevation to the top of a building is found to be 13 degrees from the ground at a distance of 4500 feet from the base of the building. Find the height of the building.
I got 2 answers for this. 385205.8259 or 1038.90686
• Dec 3rd 2006, 03:27 PM
OReilly
Quote:

2.Solve the equation http://hosted.webwork.rochester.edu/...b0c587f2f1.png by factoring.
I know that one of the solutions is definitly -8. The other one I'm getting 5/4 but it says that's wrong.

$\displaystyle 4x^2+37x+40=(x+8)(4x+5)$

x+8=0 -> x=-8
and
4x+5=0 -> x=-5/4
• Dec 3rd 2006, 03:43 PM
I figured out number 4 so I don't need help on that one.
• Dec 3rd 2006, 03:45 PM
topsquark
Quote:

1. NASA launches a rocket at http://hosted.webwork.rochester.edu/...388a5f95d1.pngseconds. Its height, in meters above sea-level, as a function of time is given by http://hosted.webwork.rochester.edu/...34bda8d651.png.
Assuming that the rocket will splash down into the ocean, at what time does splashdown occur?

The rocket splashes down after ______ seconds.

How high above sea-level does the rocket get at its peak?

The rocket peaks at _______ meters above sea-level.

Splashdown:
We are looking for a point in time where the rocket is at sea level. So we need:
$\displaystyle h(t) = -4.9t^2 + 67t + 241 = 0$

$\displaystyle t = \frac{-67 \pm \sqrt{67^2 - 4 \cdot (-4.9) \cdot 241}}{2 \cdot (-4.9)}$

I get that t = -2.95738 s or t = 16.6308 s. We may discard the negative time as unphysical.

Maximum height:
We are looking for a point in time where the vertical speed is zero (momentarily). So we set the first derivative of the height function to zero:
$\displaystyle h'(t) = -9.8t + 67 = 0$

$\displaystyle t = 6.83673 \, s$

This is WHEN the rocket reaches maximum height, so what is the maximum height?
$\displaystyle h(6.83673 \, s) = 470.031 \, m$

-Dan
• Dec 3rd 2006, 03:45 PM
I also figured out number 5
• Dec 3rd 2006, 03:47 PM
topsquark
Quote:

3.Complete the equation of the circle centered at (-1,-1) with radius 6.
I'm getting 34=0 for this but I'm not to sure what the equation for a cirlce is.

The general equation for a circle centered at (h, k) with radius r is:
$\displaystyle (x - h)^2 + (y - k)^2 = r^2$

$\displaystyle (x + 1)^2 + (y + 1)^2 = 36$

-Dan
• Dec 3rd 2006, 03:49 PM
Topsquark, It says that 6.83673seconds is wrong. The other part is right though.
• Dec 3rd 2006, 03:54 PM
Quote:

Originally Posted by topsquark
The general equation for a circle centered at (h, k) with radius r is:
$\displaystyle (x - h)^2 + (y - k)^2 = r^2$

$\displaystyle (x + 1)^2 + (y + 1)^2 = 36$

-Dan

It has to equal 0. What would it be? 2x^2+4x-32 is what i get and it's wrong.
• Dec 3rd 2006, 04:39 PM
topsquark
Quote:

Originally Posted by topsquark
Splashdown:
We are looking for a point in time where the rocket is at sea level. So we need:
$\displaystyle h(t) = -4.9t^2 + 67t + 241 = 0$

$\displaystyle t = \frac{-67 \pm \sqrt{67^2 - 4 \cdot (-4.9) \cdot 241}}{2 \cdot (-4.9)}$

I get that t = -2.95738 s or t = 16.6308 s. We may discard the negative time as unphysical.

Maximum height:
We are looking for a point in time where the vertical speed is zero (momentarily). So we set the first derivative of the height function to zero:
$\displaystyle h'(t) = -9.8t + 67 = 0$

$\displaystyle t = 6.83673 \, s$

This is WHEN the rocket reaches maximum height, so what is the maximum height?
$\displaystyle h(6.83673 \, s) = 470.031 \, m$

-Dan

Quote: