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!

Help .

  1. Sep 26, 2003 #1
    Help .......

    I need help with these problems and anyone who can respond to these asap... it would be very much apprciated... so here they are I am so confused please help me...

    1.) what is the magnitude of the average acceleration of a skier who starting from rest reaches a speed of 5.5m/s when going down a slope for 5.2 s...... part b.. how far does the skier go in this time?

    2.)A drag racer starting from rest, speeds up for 402 meters with an acceleration of +24.0m/s^2 a paracutte opens slowing the car down with and acceleration of -6.90 , how fast is the racer going 395 m after the paracutte opens?

    3.) A ball is thrown upward from a 26.6m tall building at and inital speed of 12m/s.. at the same instant a person running on the ground at a distance of 29.8m from the building.. what must be the average speed of the person if he is to catch the ball at the bottom of the building?


    ANY HINTS PLEASE.. ITS IMPORTANT.. thanks.... :frown:
     
  2. jcsd
  3. Sep 26, 2003 #2

    Janus

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    First you need to show us that you at least tried to do the problems on your own. How far did you get, and where did you get stuck?
     
  4. Sep 27, 2003 #3
    let's see...

    solution for #1.

    s = displacement
    a = acceleration
    v = final velocity
    t = time
    u = initial velocity

    Use this equation to find acceleration:

    a = v/t
    a = 1.05m/s^2

    To find the displacement:

    s = ut + 1/2at^2
    s = 1/2(1.05)(5.2)^2
    s = 14.2m

    Solution for number 2.

    Find the final velocity for the first part of the question, and use it as the inital velocity for the second part of the question.

    v^2 = u^2 + 2as
    v^2 = 2(24)(405)
    v^2 = 19440
    v = 139m/s

    Now use the same equation, just substitute the new numbers.

    v^2 = u^2 + 2as
    v^2 = 139^2 + 2(-6.9)(305)
    v^2 = 19321 - 5451
    v = 118m/s

    Solution for number 3:

    Since the time for both the runner and the ball is the same, find the time for the ball first, then use that number for the time of the runner to find the average velocity.

    s = 26.6m
    a = 9.8m/s^2
    v = -------
    t = t
    u = -12m/s

    t = -u + Square Root of u^2 - 4(1/2a)(s) / a
    t = 12 + Square Root of 144 - 4(4.9)(26.6) / 9.8
    t = 12 + Square Root of 144 - 521.4 / 9.8
    t = 31.4 / 9.8
    t = 3.2s

    To find the average velocity of the runner:

    v = d/t

    v = 29.8/3.2
    v = 9.3m/s

    Good luck
     
  5. Sep 27, 2003 #4

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    A note to both Mesmer17 and Janus: Please read "Read before posting" before posting!

    Or at least read Janus' response.
     
  6. Sep 27, 2003 #5
    Should I take down my solutions, if he is to show how far he got and where got problems? I just remember when I first started taking Physics, I didn't even know where to start, I just thought hopefully by posting the solutions, he can see HOW to do it. That way he can try doing the same questions alone, and if he gets stuck, he can always come back and look. However, I respect this board's decision that they're suspose to show work first, so just let me know, or you can take my post down yourself.
     
    Last edited by a moderator: Sep 27, 2003
  7. Sep 27, 2003 #6

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Nah, let 'em stay. If I were really uptight about it I'd have taken it down myself. I'll just content myself with pointing out to Mesmer17 that he had better check the answers himself!

    Did you notice that in part (b) of the first problem, you are assuming that the "average acceleration" is the instantaneous acceleration for the entire time?
    It doesn't work that way.

    Suppose the acceleration had been 0 for t=0 to 5.1 seconds, then was 55 m/s2 for .1 second. The final speed would be 5.5 m/s just as in the problem so the "average acceleration" would be 1.05 m/s2, just as you got.

    But his distance traveled would be (1/2)55(.1)2= 0.275 meters. Not at all what you got!
     
  8. Sep 27, 2003 #7
    Didn't they want the average acceleration (assuming to use the average acceleration to find the distance)? If not just ignore my posts, I'm stupid.
     
    Last edited by a moderator: Sep 27, 2003
  9. Sep 29, 2003 #8

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Yes, they asked for "average acceleration" in part a.

    In part b, they asked for "distance". My point was that that cannot be calculated from the "average acceleration". Indeed, the problem itself does not give enough information to answer part b.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Help .
  1. How high is the cliff? (Replies: 6)

  2. Help! ! ! ! ! (Replies: 1)

  3. Help :/ (Replies: 2)

Loading...