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Homework Help: Calculating uniform acceleration using two time intervals and its known distances

  1. Apr 16, 2012 #1
    Hi guys,

    New here, i have a question which has been doing my head in for days now. Hope someone might know how to answer it.

    The question is: A car travels with uniform acceleration from rest and is observed to cover distances of 6m and 7.5m respectively in two consecutive intervals of 1 second each.

    i) the value of the uniform acceleration,

    ii) the velocities at the beginning and end of each of these two seconds,

    iii) the total distance travelled from rest till end of the 7.5m distance period.

    I can work out part ii) and iii) if i had acceleration, but i cannot seem to work it out.
    I have drawn a velocity/time graph but it just confuses me even more.

    From my calculations i work a= 1.5m/s2

    Any help would be great!

    Thanks Bmrboi
  2. jcsd
  3. Apr 16, 2012 #2
    The acceleration is [itex]\dfrac{dv}{dt}=\dfrac{\Delta v}{\Delta t}[/itex]
  4. Apr 16, 2012 #3
    Thanks for the reply, are you meaning a= (7.5m-6m)/ (2s-1s)? Which Equals 1.5m/s2?
  5. Apr 16, 2012 #4
    Yes. The acceleration is the slope of the velocity/time graph.

    To be correct, the speed is not in m, but in m/s.

    The first time interval is 6m in 1s or 6m/1s=6m/s. This will get more important when problems become less obvious and you can recognize mistakes by checking for wrong dimensions.
  6. Apr 17, 2012 #5
    Yes you are right. I have answers now! thanks!
  7. Apr 17, 2012 #6
    No, this is not quite right. If your acceleration is constant then you can indeed use this formula, but how have you figured out that the speed at the end of the first second is 6 m/s? If you think for a second you'll realize that you have no basis for thinking that. It would only be true if it uniformly had a speed of 6 m/s for the first second, in which case it would have zero acceleration (which isn't true). You can also prove that 1.5 m/s[itex]^2[/itex] is wrong: that means that every second, the speed grows by 1.5 m/s, by definition. But your claim is that the speed is 6m/s after the end of one second. This is a contradiction.

    So you'll have to try again, with a different kinematics relationship.
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