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

Determine distance with constant acceleration

  1. Jul 8, 2011 #1
    1. The problem statement, all variables and given/known data
    A particle moving at 5m/s reverses its direction in 1 s to move at 5m/s in the opposite direction. If its acceleration is constant, what distance does it travel?

    2. Relevant equations

    x=vot+0.5at^2; a=(vf-vi)/t

    3. The attempt at a solution

    total time is 1s therefore acceleration is 10m/s^2 and then i'm stuck because when I input the value into the equation I'm not getting the answer of 2.5m which is the correct answer.
  2. jcsd
  3. Jul 8, 2011 #2


    User Avatar
    Science Advisor
    Homework Helper

    hi brunettegurl! :wink:

    i think it means that it goes 1.25m in 0.5s, and then 1.25m back again, = 2.5m

    try again :smile:
  4. Jul 8, 2011 #3
    That's slightly tricky because in the equation you gave x is not the distance traveled but merely a coordinate describing the location. To highlight the difference: if I go for a stroll and come back home then my location will be the same as before, but I obviously traveled some distance. In your case, if you properly plugged in all values (taking into account the relative signs, because the acceleration is in opposite direction than the starting velocity) you should have ended up with x=0, which is completely correct but not the sought-for variable.
    The solution to your problem is to realize that the distance traveled consists of a part that the particle moved to positive x-direction and a part where it traveled into negative x-directions. Treat these two parts separately and you get the correct result.

    btw.: In the future, please also give the incorrect result you got, because that can greatly help understanding what went wrong.
  5. Jul 8, 2011 #4

    I'm still a little confused my book states that to find distance we use half the total time in the equation w/accleration and then solve for x by doubling the answer for total distance by doing this ::

    a=(5-(-5))/0.5= 20m/s^2

    and when plugged into the equation x=(-5)(1)+ (0.50)(20)(1)^2
    x(displacemnt)=5m and doubling this answer gives 10m

    tiny-tim how did you get 1.25 as the distance travelled in 0.5s??

  6. Jul 8, 2011 #5


    User Avatar
    Science Advisor
    Homework Helper

    hi brunettegurl! :smile:

    no, it's not 20 m/s2, it's still 10, isn't it? :wink:
  7. Jul 8, 2011 #6
    Im sorry how is it still 10 when you divide by time which is 0.5 now according to the explanation (1/2 the time) making the 10 now a 20 as acceleration
  8. Jul 8, 2011 #7


    User Avatar
    Science Advisor
    Homework Helper

    it goes from +5 to -5 in 1 s (or from +5 to 0 in 0.5 s) …

    that's 10 !
  9. Jul 8, 2011 #8
    ohhh so for acceleration the vf would be 0 and vi would be +5 giving the acceleration of 10 .. correct?

    so then when we sub in the values in x=(5)(1)+ (0.50)(10)(1)^2
    i still get an x(displacemnt) value of =10m and doubling this answer gives 20m

    do i continue to use 0.5 as my time value and then since it is reversing in the same scalar value of 5 just dbl the new x-value??
  10. Jul 8, 2011 #9


    User Avatar
    Science Advisor
    Homework Helper

    (try using the X2 icon just above the Reply box :wink:)

    that 1 should be 0.5, and that + should be - :redface:
  11. Jul 8, 2011 #10
    Ok Thanks I understand now ... I think what confused me when I was reading the explanation in my book was the implication that 2 different time values were to be used... Thank you :)
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook