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!

Kinematic -- If object in 5th and 6th sec passes 20

  1. Jul 20, 2014 #1
    1. The problem statement, all variables and given/known data
    Object from state of inaction starts to move with equal acceleration. If objec in 5th and 6th sec passes 20, whats acceleration.


    2. Relevant equations
    s=v0*t+a*t*t/2 v0=0


    3. The attempt at a solution
    s=a*t*t/2
    a=s/2
    a=10
     
  2. jcsd
  3. Jul 20, 2014 #2

    Nathanael

    User Avatar
    Homework Helper

    You can't use "v0=0" because at "v0" the object has already been accelerating for 5 seconds.
     
  4. Jul 20, 2014 #3
    Does the object covered 20m distance between 5sec. and 6sec. of the journey?
    During the start of journey
    u=0.Let the acceleration be a then after 5 sec. the velocity of the object=5a(using v=u+at)
    For the motion of object between 5sec. and 6sec.---
    u=5a,S=20m(given),t=1sec.
    Using S=ut+0.5at^2 Equation
    20=5a+0.5a.Find the value of a to get the answer.
     
  5. Jul 21, 2014 #4
    In case you want a quick summary on the topic on Kinematics, here's a good resource:

     
    Last edited by a moderator: Sep 25, 2014
  6. Jul 21, 2014 #5

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    I don't know what you mean by "at v0". At t= 0, the object is just starting to accelerate with initial speed v0= 0 so this formula can be used to find the distances at t= 5 and t= 6.
     
  7. Jul 21, 2014 #6

    Nathanael

    User Avatar
    Homework Helper

    By "at v0" I meant the "v0" that the OP was using in his equations. (I wanted to keep it as simple as possible because he didn't seem to be a native english speaker.)

    If you look at his equations then you can see he assumed "(Δ)t=1" and "s=20" which means he created his equation to describe the time interval from t=5 to t=6.

    That is why I said he can't use v0=0, because HIS "v0" is not 0


    (You can certainly solve the problem using v0=0 but my comment was on the method that he attacked the problem with.)
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Kinematic -- If object in 5th and 6th sec passes 20
  1. Two Object Kinematics (Replies: 2)

Loading...