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    Hi guys,

    I am in real need of some help of this question!

    A particle travels in straight line with uniform acceleration. The particle passes through 3 points, A, B and C lying in that order at times t=0, t=2s, t=5s. If BC=30meters, and the speed of the particle when at B is 7ms^-1 find the acceleration of the particle and its speed when at A.

    Okay so I am guessing it is a simultaneous equation right?

    So yeah I am pretty much stuck aha.

    From AB; S=x
    U=u
    V=doesnt matter
    A=a
    t=2s

    From AC S=30+x
    U=7ms^-1
    V= doesnt matter
    A=a
    T= 7s

    I really have no clue what to do from here now as I can't solve it!
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    (Original post by J DOT A)
    Hi guys,

    I am in real need of some help of this question!

    A particle travels in straight line with uniform acceleration. The particle passes through 3 points, A, B and C lying in that order at times t=0, t=2s, t=5s. If BC=30meters, and the speed of the particle when at B is 7ms^-1 find the acceleration of the particle and its speed when at A.

    Okay so I am guessing it is a simultaneous equation right?

    So yeah I am pretty much stuck aha.

    From AB; S=x
    U=u
    V=doesnt matter
    A=a
    t=2s

    From AC S=30+x
    U=7ms^-1
    V= doesnt matter
    A=a
    T= 7s

    I really have no clue what to do from here now as I can't solve it!
    Use s=ut+1/2at^2 for BC to get the acceleration of the particle and then use v=u+at for AB to find the initial speed (u).
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    (Original post by Gemini92)
    Use s=ut+1/2at^2 for BC to get the acceleration of the particle and then use v=u+at for AB to find the initial speed (u).
    Oh right so is the acceleration constant throughout process? If I use from BC I get acceleration as -3/5 ms^2. The answer is 2ms^2?
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    (Original post by J DOT A)
    Oh right so is the acceleration constant throughout process? If I use from BC I get acceleration as -3/5 ms^2. The answer is 2ms^2?
    Yes, the acceleration is constant, that's what's meant by uniform acceleration. I got the acceleration to be 2ms^-2. What values are you using to work out the acceleration?
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    (Original post by Gemini92)
    Yes, the acceleration is constant, that's what's meant by uniform acceleration. I got the acceleration to be 2ms^-2. What values are you using to work out the acceleration?

    LOL my bad, sorry:/
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    The answer is 3m/s.
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    U(a)=x (unknown)
    V(a)=U(b)=7
    BC=30
    from v=u+at
    7=x+2a

    then

    s=U(b)t+0.5at^2
    30=21+4.5a
    a=2

    then U(a)=7-4
    U(a)=3 m/s
 
 
 
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