1. Limited time only! Sign up for a free 30min personal 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!

Homework Help: Concerning motion

  1. Mar 6, 2005 #1
    Hi, I was given the following question:

    Q: A body starts moving with an acceleration [itex]a=const[/itex]. After a time [itex]t[/itex] it has covered the distance [tex]\Delta x[/itex]. What is its speed at time [itex]t[/itex]?

    Answer: [tex]\frac{2\Delta x}{t}[/tex]

    Can somebody say how to arrive at this answer (both mathematically and intuitionally)? I know that [itex]\Delta x/t[/tex] would be the speed w/o acceleration, so the "2" must come from acceleration, but where does it come from?

    Can the answer be obtained from the formula [itex]v=v_0+at^2[/itex]?

    - Kamataat
  2. jcsd
  3. Mar 6, 2005 #2
    [tex]x-x_0=\frac{1}{2}(v_0+v_f)\Delta{t},\ \ v_0=0 m/s[/tex]
  4. Mar 6, 2005 #3
    Thank you!

    - Kamataat
  5. Mar 6, 2005 #4


    User Avatar
    Science Advisor

    Another way: for CONSTANT acceleration, the "average" speed is just the average of the starting speed and ending speed: [itex]\frac{v_i+v_f}{2}[/itex]. Since the starting speed is 0, if the ending speed is v, this is [itex]\frac{v}{2}[/itex] so the distance covered in time [itex]\Delta t[/itex] would be [itex]\Delta x =\frac{v}{2}\Delta t}[/itex] so [itex]v= 2\frac{\Delta x}{\Delta t}[/itex].

    CAUTION: This is only true for CONSTANT acceleration!
    Last edited by a moderator: Mar 6, 2005
  6. Mar 7, 2005 #5
    Halls, that's what my equation says :) [itex]1/2(v_0 + v_f) =[/itex] average velocity :smile:
  7. Mar 7, 2005 #6
    yup, i get it now. tnx
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook