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    When the cyclist reaches the top of the hill, her speed is 3 m s-1. She subsequently acceleratesuniformly so that in the fifth second after she has reached the top of the hill, she travels 12 m.

    (b) Find her speed at the end of the fifth second.


    This is so badly worded i'm so confused. it says that IN the 5th second she travels 12m, doesnt this imply she is travelling at 12m/s???? but its asking me to find speed at the end of the 5th second????

    im so confused someone help
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    (Original post by dnr_23)
    This is so badly worded i'm so confused. it says that IN the 5th second she travels 12m, doesnt this imply she is travelling at 12m/s???? but its asking me to find speed at the end of the 5th second????

    im so confused someone help
    NO. It would be true if she was travelling at constant speed, but that's not the case here.

    Wording of the question is fine, IMO.
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    i get you, it is badly worded. I had a similar problem in another solomon paper.
    But just use u=3, t=5,s=12 and v=? and solve using suvat.
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    In the 5th second i.e.in the time between seconds 4 and 5, the distance travelled is 12m. Let a = acceleration, u = speed at the end of the 4th second.

    So the distance travelled in the fifth second = 12 = u + 0.5at^2 = u + 0.5a

    Knowing that the acceleration is constant, u = initial speed + at (where t=4) i.e. u = 3 + 4a

    So, substitute this into the first equation to get 12 = (3 + 4a) + 0.5a = 3 + 4.5a so 4.5a = 9 and a = 2 m/s^2
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    thank u everyone!!
 
 
 
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