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Kinematic question, i think it's hard

  1. May 3, 2008 #1
    kinematic question, i think it's hard:(

    1. The problem statement, all variables and given/known data
    Two camels leave the spinx at time 52.5 s apart. The first moves off with a uniform acceleration of 1.30 m/s2, which it maintains for 1.92 s. It then continues with the acquired speed. The second camel moves of with a uniform acceleration of 1.45 m/s2, which it maintains for 2.84 s. It then contiues with the acquired speed. Claculate:
    a) the distance they have both travelled when the second camel overtakes the first camel, which has walked in a straight line
    b) the time taken by the second camel to catch the first.

    2. Relevant equations
    a= v-u/t?? not sure..

    3. The attempt at a solution

    I really don't know how to put the 'two camels leave spinx at 52.5 seconds apart, i dont understand the 52.5 s apart and where to put that into equation. I'm pretty styumped with this question, could someone pls help? I have acceleration for both camels and the time but thats all i have. does u=0? Could i then use the equation s=ut+0.5at2 for each camel to find out their distances and then add them together. I just don't get the 52.5 seconds apart:(
    and with question b) i know that the second camel has a greater acceleration than the first but other than that, i dont get it:(
    Last edited: May 3, 2008
  2. jcsd
  3. May 3, 2008 #2
    Camel A leaves at time t=0
    Camel B leaves at time t=52.5
    Camel A acceleration = 1.30
    Camel A speed for 1.92 seconds=[tex]1.30t[/tex]

    Camel B acceleration = 1.45

    Camel B speed for 2.84 seconds=[tex]1.45t[/tex]

    Now, position is [tex]\frac{1}{2}(v(t))^2[/tex], which is derived by integration.
    Position of the camel A at t=1.92 = [tex]\frac{1}{2}((1.92)(1.30))^2[/tex]

    Velocity of camel A at t = 1.92 = [tex](1.92)(1.30)[/tex]
    Position of camel B at t = 55.34 = [tex]\frac{1}{2}((2.84)(1.45))^2[/tex]

    Velocity of camel B at t = 55.34 = [tex](2.84)(1.45)[/tex]

    And then you solve the problem. Which should be pretty easy.
  4. May 3, 2008 #3
    For uniform acceleration:

    v = at + vi, in which vi is initial velocity (which is 0m/s)
    x = (1/2)*a*t^2 + vi*t + xi, in which vi is initial velocity (0m/s) and xi is initial displacement (which is 0m).

    Since the second camel left 52.5s after the first camel, it starts 52.5s behind the first camel, meaning the time for the second camel is (t - 52.5)

    So, the 2 equations you need for the first camel are:
    v = at
    x = (1/2)*a*t^2
    and the 2 you need for the second camel are:
    v = a(t - 52.5)
    x = (1/2)*a*(t - 52.5)^2
  5. May 3, 2008 #4
    Why don't you guys use LATEX?


    So here you have two ways of looking at the problem =D
  6. May 4, 2008 #5
    52.5 seconds apart means if one camel started of at t=0s the other camle started off att=52.5 s there's a diff of this much time in between them which wud obviously result in diff in distance covered..just calculate the distance travveled by both of them...at the pt where the distance travelled by 2nd camel just becomes equal to that travelled by 1st camel.is the pt where the 2nd one overtakes..use equations of motion in a straight line
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