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    A car is travelling along a straight horizontal road. The car takes 120 s to travel between two sets of traffic lights which are 2145 m apart. The car starts from rest at the first set of traffic lights and moves with constant acceleration for 30 s until its speed is 22 m s–1. The car maintains this speed for T seconds. The car then moves with constant deceleration, coming to rest at the second set of traffic lights. (a) Sketch, in the space below, a speed-time graph for the motion of the car between the two sets of traffic lights. (2)

    (b) Find the value of T. (3) A motorcycle leaves the first set of traffic lights 10 s after the car has left the first set of traffic lights. The motorcycle moves from rest with constant acceleration, a m s–2, and passes the car at the point A which is 990 m from the first set of traffic lights. When the motorcycle passes the car, the car is moving with speed 22 m s–1.
    (c) Find the time it takes for the motorcycle to move from the first set of traffic lights to the point A. (4)
    (d) Find the value of a.

    Hi, could someone help with with part d please? I have done abc and found that it takes 50s for the motorcyle to catch up with the car. However for part d , why can't do you the equation a=(v-u)/2 ? The equation that the mark scheme has included t and d. But surely a=(v-u)/t would give me the same thing since both are used to find a. Is this to do with a being constant? Thanks.
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    (Original post by coconut64)
    A car is travelling along a straight horizontal road. The car takes 120 s to travel between two sets of traffic lights which are 2145 m apart. The car starts from rest at the first set of traffic lights and moves with constant acceleration for 30 s until its speed is 22 m s–1. The car maintains this speed for T seconds. The car then moves with constant deceleration, coming to rest at the second set of traffic lights. (a) Sketch, in the space below, a speed-time graph for the motion of the car between the two sets of traffic lights. (2)

    (b) Find the value of T. (3) A motorcycle leaves the first set of traffic lights 10 s after the car has left the first set of traffic lights. The motorcycle moves from rest with constant acceleration, a m s–2, and passes the car at the point A which is 990 m from the first set of traffic lights. When the motorcycle passes the car, the car is moving with speed 22 m s–1.
    (c) Find the time it takes for the motorcycle to move from the first set of traffic lights to the point A. (4)
    (d) Find the value of a.

    Hi, could someone help with with part d please? I have done abc and found that it takes 50s for the motorcyle to catch up with the car. However for part d , why can't do you the equation a=(v-u)/2 ? The equation that the mark scheme has included t and d. But surely a=(v-u)/2 would give me the same thing since both are used to find a. Is this to do with a being constant? Thanks.
    That equation doesn't have a in it though :/
    isn't the correct version of that s=\left(\dfrac{u+v}{2}\right)t?
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    I don't think that equation exists- do you mean a=v-u/t instead of 2?
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    (Original post by Tsunaomi)
    I don't think that equation exists- do you mean a=v-u/t instead of 2?
    Oh yeah, sorry that's the one I was talking about a=(v-u)/t. Thanks.
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    (Original post by Steelmeat)
    That equation doesn't have a in it though :/
    isn't the correct version of that s=\left(\dfrac{u+v}{2}\right)t?
    Sorry I meant a=(v-u)/t . Thanks.
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    (Original post by coconut64)
    Oh yeah, sorry that's the one I was talking about a=(v-u)/t. Thanks.
    No problem, I always used to get 2 and T mixed up too, they sound so similar in my head lol
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    (Original post by Tsunaomi)
    No problem, I always used to get 2 and T mixed up too, they sound so similar in my head lol
    Yh hopefully this wouldn't happen in the exam. Anyway any idea why I can't use that equation to work out a ??
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    (Original post by Steelmeat)
    That equation doesn't have a in it though :/
    isn't the correct version of that s=\left(\dfrac{u+v}{2}\right)t?
    this equation does exist and its right but we hardly use it in M1
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    (Original post by jovdawesome)
    this equation does exist and its right but we hardly use it in M1
    hardly yes.... but its still there... and i'd rather use it than s=ut+\frac{1}{2} at^2

    and v^2 = u^2 +2as
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    (Original post by Steelmeat)
    hardly yes.... but its still there... and i'd rather use it than s=ut+\frac{1}{2} at^2

    and v^2 = u^2 +2as
    Hi for this question I posted why can't I use a=(v-u)/t to find a? Thanks
 
 
 
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