Results 1 to 9 of 9

Math Help - acceleration motion problem.

  1. #1
    Junior Member
    Joined
    Nov 2006
    Posts
    73

    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 seconds. Its height, in meters above sea-level, as a function of time is given by .
    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 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
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member OReilly's Avatar
    Joined
    Mar 2006
    Posts
    340
    Quote Originally Posted by badandy328 View Post

    2.Solve the equation 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.
    4x^2+37x+40=(x+8)(4x+5)

    Correct answer is:
    x+8=0 -> x=-8
    and
    4x+5=0 -> x=-5/4
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Nov 2006
    Posts
    73
    I figured out number 4 so I don't need help on that one.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    10,059
    Thanks
    369
    Awards
    1
    Quote Originally Posted by badandy328 View Post
    1. NASA launches a rocket at seconds. Its height, in meters above sea-level, as a function of time is given by .
    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:
    h(t) = -4.9t^2 + 67t + 241 = 0

    Using the quadratic formula:
    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:
    h'(t) = -9.8t + 67 = 0

    t = 6.83673 \, s

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

    -Dan
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Nov 2006
    Posts
    73
    I also figured out number 5
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    10,059
    Thanks
    369
    Awards
    1
    Quote Originally Posted by badandy328 View Post
    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:
    (x - h)^2 + (y - k)^2 = r^2

    So your circle will be:
    (x + 1)^2 + (y + 1)^2 = 36

    -Dan
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Junior Member
    Joined
    Nov 2006
    Posts
    73
    Topsquark, It says that 6.83673seconds is wrong. The other part is right though.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Junior Member
    Joined
    Nov 2006
    Posts
    73
    Quote Originally Posted by topsquark View Post
    The general equation for a circle centered at (h, k) with radius r is:
    (x - h)^2 + (y - k)^2 = r^2

    So your circle will be:
    (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.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    10,059
    Thanks
    369
    Awards
    1
    Quote Originally Posted by topsquark View Post
    Splashdown:
    We are looking for a point in time where the rocket is at sea level. So we need:
    h(t) = -4.9t^2 + 67t + 241 = 0

    Using the quadratic formula:
    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:
    h'(t) = -9.8t + 67 = 0

    t = 6.83673 \, s

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

    -Dan
    Quote Originally Posted by badandy328 View Post
    Topsquark, It says that 6.83673seconds is wrong. The other part is right though.
    That's because there is no answer t = 6.83673 s. It should be 16.6308 s.

    -Dan
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Motion with Uniform Acceleration
    Posted in the Math Topics Forum
    Replies: 3
    Last Post: October 4th 2009, 05:08 AM
  2. Motion with uniform acceleration??
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: September 16th 2009, 11:39 AM
  3. Acceleration Problem
    Posted in the Calculus Forum
    Replies: 4
    Last Post: May 10th 2008, 12:26 PM
  4. Replies: 4
    Last Post: February 10th 2008, 05:30 PM
  5. Replies: 2
    Last Post: November 29th 2007, 03:12 AM

Search Tags


/mathhelpforum @mathhelpforum