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    For my brother, he's studying his A Levels -

    A cyclist is travelling at a constant speed of 3 ms-1 as he starts to roll down a hill. He rolls down the hill with a constant acceleration. During the third second, he travels at a distance of 6 m.

    a) Calculate the cyclist's acceleration.
    b) Calculate how far he travels during the 4th second.

    I would appreciate explanations of how you understood and got the answer
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    bumpity
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    acceleration = v-u/t


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    Use suvat equations
    S = displacement / distance
    U = initial velocity
    V = final velocity
    A = acceleration
    T = time

    V = u+at


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    (Original post by AdeptDz)
    acceleration = v-u/t


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    Can I please ask if you could directly give a solution to the answer?
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    I got a = 0, But im not sure i jsut started learning this too.
    I assumed final velocity was 0 as it doesn;t explicitly say it but "constant speed of 3ms-1" means he didnt slow down so im guessing he was going the same speed all the way up to 3 seconds so same velocity, but ill do question b now
    a=v-u/t
    a=3-3/3 = 0

    I got b is 9 or 12, realistically i got 12, but i probably made a mistake somewhere so hopefully someone comes and clears this up
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    What did you get?
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    Basically for a) all you do is use the suvat equations and solve a simultaneous equation using the distance at 2 seconds (s) and the distance at 3 seconds (s+6):
    u = 3ms-1
    a = constant
    t = 2/3 seconds
    s = distance (varied)

    Equation used: s = ut + 0.5(at2)
    Simultaneous equations:
    s = 3(2) + 0.5a(2)2
    s+6 = 3(3) + 0.5a(3)2

    When I solved it I got the acceleration as 1.2ms-2

    Once a) was done b) is simple as the acceleration is known.
    Distance for the 4th second (t=3 to t=4)
    when t=3,
    s = ut + 0.5(at2)
    s = 3(3) + 0.5(1.2)(3)2 = 14.4m
    when t=4,
    s = 3(4) + 0.5(1.2)(4)2 = 21.6m

    So the distance travelled in the 4th second is 21.6-14.4 = 7.2m

    (btw I just did as physics so please correct me if I made a mistake)
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    (Original post by LogicIsKey)
    Basically for a) all you do is use the suvat equations and solve a simultaneous equation using the distance at 2 seconds (s) and the distance at 3 seconds (s+6):
    u = 3ms-1
    a = constant
    t = 2/3 seconds
    s = distance (varied)

    Equation used: s = ut + 0.5(at2)
    Simultaneous equations:
    s = 3(2) + 0.5a(2)2
    s+6 = 3(3) + 0.5a(3)2

    When I solved it I got the acceleration as 1.2ms-2

    Once a) was done b) is simple as the acceleration is known.
    Distance for the 4th second (t=3 to t=4)
    when t=3,
    s = ut + 0.5(at2)
    s = 3(3) + 0.5(1.2)(3)2 = 14.4m
    when t=4,
    s = 3(4) + 0.5(1.2)(4)2 = 21.6m

    So the distance travelled in the 4th second is 21.6-14.4 = 7.2m

    (btw I just did as physics so please correct me if I made a mistake)
    Why did you use the time at 2 seconds? Don't think we've done that yet maybe I'll learn next lesson


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