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Kinematics - find displacement between A and B watch

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    1. The problem statement, all variables and given/known data

    A cheetah is estimated to be able to run at a maximum speed of 100km.h^1 whilst and antelope can run at a maximum speed of 65km.h^1. A cheetah at rest sees an antelope at rest and starts running towards it. The antelope immediately starts running away. Both cheetah and antelope are assumed to move with constant acceleration and reach their maximum speeds in 4 seconds. Assuming that both run along the same straight line and that the cheetah catches the antelope in 15 seconds, find the distance between the animals when they first started moving.


    2. Equations I have been given
    v=u+at
    s=ut+1/2at^2
    v^2=u^2+2as
    s=1/2(u+v)t

    I have been staring at this for a good hour now and am yet to come up with the answer which I know to be 126m. How do i find this value using those equations and the data given? Our mechanics tutor is away with an illness and we have been set work which we haven't been taught properly so any help would be greatly appreciated!

    here is what i have gotten so far but the final answer is not 126m which i know to be correct, any ideas?


    cheetah-v=250/9 u=0 t=4 a=?

    250/9 =4a
    (250/9)/4=125/18
    125/18=a

    antelope-v=325/18 u=0 t=4 a=?

    (325/18)/4=a
    325/72=a

    cheetah displacement -

    first 4 secs
    s=ut+1/2at^2
    s=0x4+1/2x125/18x4^2
    s=500/9

    final 11 secs
    s=1/2(u+v)t
    1/2(0+250/9)11
    s=1375/9

    total S = 625/3


    antelope displacement -

    first 4 secs
    s=ut+1/2at^2
    s=0x4+1/2x325/72x4^2
    s=325/9

    final 11 secs
    s=1/2(u+v)t
    1/2(0+325/18)11
    s=3575/36

    total S = 1625/12

    s=(625/3)-(1625/12)=875/12

    Does that look correct? According to the answer i know to be correct (its on the back of the assignment paper) it is wrong :s

    Thanks again for this!
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    100km/h = 100*1000/60*60 = 250/9 m/s
    65km/h = 65*1000/60*60 = 650/36 m/s

    Then, simply work out the distance travelled by the antelope in the 15 seconds (do this in two parts, the 0>4 seconds, then 4>15 at constant velocity).
    Do exactly the same thing for the Cheetah. The excess distance the Cheetah travels over the antelope will be the distance between the two at the start.
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    Cheetah (between 1-4 seconds)
    S= S
    U=0
    V= 250/9
    A =
    T = 4

    S= 250/18 * 4 = 500/9 m

    Between (4-15 seconds)
    S=S
    U=250/9
    V=U
    A=
    T=11

    S = 250/9 * 11 = 2750/9 m

    TOTAL = 3250/9 m


    Do the same for the antelope.
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    (Original post by Noble.)
    Cheetah (between 1-4 seconds)
    S= S
    U=0
    V= 250/9
    A =
    T = 4

    S= 250/18 * 4 = 500/9 m
    dont you mean 250/9*4 ?
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    (Original post by derRoboter)
    dont you mean 250/9*4 ?
    No, it's U+V/2 * T

    U=0, so it's simple V/2 * T
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    (Original post by Noble.)
    No, it's U+V/2 * T

    U=0, so it's simple V/2 * T
    dont you need to work out the acceleration for the first 4 seconds first?

    or is that just overcomplicating it?
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    1) 100 km/h = (100*1000)/3600 m/s
    65 km/h = (65*1000)/3600 m/s

    2) If they accelerate at a constant rate for 4 seconds, then their average speed in this time is half their max speed, i.e half the values you have just worked out. Half their max speed x 4 seconds is hence the distance travelled in this time.

    3) Then times their speeds by 11 seconds.

    4) Add this to the 4 second distance.

    5) The amount by which the cheetah's distance is greater is the distance between them at the start.

    Ignore SUVAT equations.
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    (Original post by Thrug)
    1) 100 km/h = (100*1000)/3600 m/s
    65 km/h = (65*1000)/3600 m/s

    2) If they accelerate at a constant rate for 4 seconds, then their average speed in this time is half their max speed, i.e half the values you have just worked out. Half their max speed x 4 seconds is hence the distance travelled in this time.

    3) Then times their speeds by 15 seconds.

    4) Add this to the 4 second distance.

    5) The amount by which the cheetah's distance is greater is the distance between them at the start.

    Ignore SUVAT equations.
    the answer that gives is not 126m, which is what the paper says the answer is supposed to be. that gives 5425/36. any ideas?
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    (Original post by derRoboter)
    the answer that gives is not 126m, which is what the paper says the answer is supposed to be. that gives 5425/36. any ideas?
    Hmm doing my method gave me the answer of 165m.
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    (Original post by Thrug)
    3) Then times their speeds by 15 seconds.

    .
    Should be 15-4=11 seconds
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    (Original post by ghostwalker)
    Should be 15-4=11 seconds
    agreed, but that gives an incorrect answer also. :/ im glad 5 of us cant get the answer lol
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    (Original post by ghostwalker)
    Should be 15-4=11 seconds
    Ah didn't read the question carefully enough, thought it was 4 then 15 :/

    Now I get the answer of 126m
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    (Original post by derRoboter)
    agreed, but that gives an incorrect answer also. :/ im glad 5 of us cant get the answer lol
    Shouldn't do - check your calculations again.
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    (Original post by derRoboter)
    im glad 5 of us cant get the answer lol
    Speak for yourself, I posted the answer on your other thread (but not the working)!

    It's easier to work this through algebraically and then put in the figures, and you'll get the formula I posted on your other thread many hours ago.
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    (Original post by ghostwalker)
    Speak for yourself, I posted the answer on your other thread (but not the working)!

    It's easier to work this through algebraically and then put in the figures, and you'll get the formula I posted on your other thread many hours ago.
    alright man, dont be like that. Ill go take a look. Thank you for helping me on this.

    + rep.

    derRoboter
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    (Original post by derRoboter)
    alright man, dont be like that. Ill go take a look. Thank you for helping me on this.

    + rep.

    derRoboter
    :p: :lol: :cheers:
 
 
 
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