Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: More uniform acceleration

  1. Feb 10, 2008 #1
    I'm having some trouble with a couple more problems. Any help is appreciated.

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

    (a) An indestructible bullet 1.90 cm long is fired straight through a board that is 10.3 cm thick. The bullet strikes the board with a speed of 426 m/s and emerges with a speed of 278 m/s. What is the total time that the bullet is in contact with the board?

    (b) A woman is reported to have fallen 34 ft from the top of a building, landing on a metal ventilator box, which she crushed to a depth of 25 in. Calculate her average acceleration while in contact with the box.

    2. Relevant equations

    (i) [tex]\Delta x = v_{o}t + \frac{1}{2}at^{2}[/tex]

    (ii) [tex]a = \frac{v^{2} - v^{2}_{o}}{2\Delta y}[/tex]

    3. The attempt at a solution

    (a) I actually stumbled upon the answer for this one, but I ended up with two different values for [tex]t[/tex] and I'm wondering what the incorrect one means exactly.

    For this problem, I first noted that the bullet is in contact with the board for both the thickness of the board (10.3 cm) as well as for the length of the bullet at both ends of the board as it exits and enters (2 * 1.90 cm). I had previously calculated the average acceleration of the bullet as it passed through the board and used that value along with this displacement and the given velocity and plugged them into equation (i) to find the time taken to pass through the board. Thus,

    [tex]v_{o}[/tex] = 426 m/s, [tex]\Delta x_{}[/tex] = [10.3 + (1.90)(2)]/100 = 0.122 m, [tex]a[/tex] = -505786.4 m/s[tex]^{2}[/tex] (calculated and verified previously)

    Plugging these into equation (i) and solving for [tex]t[/tex] using the quadratic formula gave me [tex]t[/tex] = 0.000366 s, 0.00132 s. The first turned out to be the correct answer, but now I'm confused as to what the second represents. Is there an easier way for calculating this time that I'm not seeing?

    (b) I had previously calculated the woman's velocity just before hitting the box and used that with the displacement of 25 in. to calculate the average acceleration using equation (ii) as such:

    [tex]v_{o}[/tex] = 46.76 ft./s (calculated and verified previously), [tex]\Delta y[/tex] = 25 in. * (1 ft./12 in.) = 2.083 ft., [tex]v_{f}[/tex] = 0 ft./s

    [tex]a[/tex] = [(0)[tex]^{2}[/tex] - (46.76)[tex]^{2}[/tex]]/(2 * 2.083) = -524.8 ft./s[tex]^{2}[/tex]

    Where have I gone wrong? Thanks.
    Last edited: Feb 10, 2008
  2. jcsd
  3. Feb 10, 2008 #2


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    For (a) I would have used the formula:

    [tex] s = \left(\frac{u+v}{2}\right) t [/tex]

    For (b) I can't really see anything wrong.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook