1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Final velocity given acceleration and initial velocity

  1. Jan 25, 2011 #1
    1. The problem statement, all variables and given/known data
    A dynamite blast at a quarry launches a chunk of rock straight upward, and 2.1 s later it is rising at a speed of 19 m/s. Assuming air resistance has no effect on the rock, calculate its speed at (a) at launch and (b) 5.2 s after the launch.


    2. Relevant equations
    v = v0 + at (I'm assuming is the only relevant one, although I'll post two others in case they are needed)
    x = x0 + v0t + a/2 t^2
    v^2 = v0*2 + 2a(x-x0)


    3. The attempt at a solution

    I solved part A by plugging into the equation v = v0 + at
    19 m/s = v0 + (-9.8 m/s^2)(2.1 s)
    And I found that the initial velocity equals 40 m/s.

    So, to solve part b, I should just have to plug the initial in and find the final. I tried that:
    v = 40 m/s + (-9.8 m/s^2)(5.2 s)
    v = -11 m/s

    However, when I entered that solution in for the homework, I was told it was wrong. I'm not really sure how to go about doing the problem if that's incorrect. I thought maybe I could have an error in rounding with significant figures.

    Thanks for the help!
     
  2. jcsd
  3. Jan 25, 2011 #2

    gneill

    User Avatar

    Staff: Mentor

    It could be that since part b asked for the speed, not the velocity, that the sign of the number reported should have been positive.
     
  4. Jan 25, 2011 #3
    The problem asked for the speed of the rock - you gave its velocity. Remember - speed is a scalar (independent of direction), and velocity is a vector (depends upon direction).


    edit: gneill and I were posting at the same time, apparently.
     
  5. Jan 25, 2011 #4
    I tried that as the answer (11 m/s instead of -11 m/s) and it was still incorrect.
     
  6. Jan 25, 2011 #5
    No, that is the correct answer. 11.38 m/s to be "precise," but you're only given two significant figures in your initial conditions. I've done it three ways, and it comes to 11.38 m/s each time. And 40 m/s is correct for a. Just now, I did it a 4th way, and I got 10.96 m/s - which still rounds to 11 m/s.
     
  7. Jan 25, 2011 #6
    Okay, so I did do the math correctly. Thanks for your help :) I'm going to email my professor and see what the issue is with that problem.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?