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    A car is approaching traffic lights at 15m/s when the driver applies the brake and comes to a stop in 45m. Find the deceleration, assumed constant, and the time taken to stop.

    Using the distance speed formula, I did 45/15 and got 3 seconds for time taken to stop, but the answer is 6 seconds. Then I did 0-15/3 = -5ms, which is 5ms for deceleration, but the answer was 2.5ms. I would have got this right if I had got 6 seconds for the other answer.
    So does anyone know what I'm doing wrong?
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    (Original post by ℓove)
    A car is approaching traffic lights at 15m/s when the driver applies the brake and comes to a stop in 45m. Find the deceleration, assumed constant, and the time taken to stop.

    Using the distance speed formula, I did 45/15 and got 3 seconds for time taken to stop, but the answer is 6 seconds. Then I did 0-15/3 = -5ms, which is 5ms for deceleration, but the answer was 2.5ms. I would have got this right if I had got 6 seconds for the other answer.
    So does anyone know what I'm doing wrong?
    Why not just use v^2=u^2+2as for the deceleration?

    Then for time it's just s=\frac{1}{2}(u+v)t

    I'm not sure where you got 45/15 from, perhaps you forgot the half in the equation? Not sure what you did for the deceleration.
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    (Original post by ℓove)
    A car is approaching traffic lights at 15m/s when the driver applies the brake and comes to a stop in 45m. Find the deceleration, assumed constant, and the time taken to stop.

    (A) Using the distance speed formula, I did 45/15 and got 3 seconds for time taken to stop, but the answer is 6 seconds. Then I did 0-15/3 = -5ms, which is 5ms for deceleration, but the answer was 2.5ms. I would have got this right if I had got 6 seconds for the other answer.
    So does anyone know what I'm doing wrong?
    You can't do a DVT calculation if there is any type of acceleration/deceleration involved.

    You would have to use a kinematics equation: I would suggest you list all the data given to you in the question and decide which equation is the best fit. So it gives you initial velocity(u) as15ms^-1 and it also tells you that it comes to rest so that means that the final velocity(v) is 0ms^-1, and the displacement is given as 45m. So the equation used should be v^2=u^2+2as and you can then rearrange you get the deceleration(a).

    (B) After arriving at the deceleration you could use another kinematics equation and again rearrange to get t. I'll let you decide yourself which equation to use.


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    (Original post by RDKGames)

    I'm not sure where you got 45/15 from, perhaps you forgot the half in the equation? Not sure what you did for the deceleration.
    I think she tried to use t=d/v to calculate the time and then put it into a=(v-u)/t. But that can't be done when there is an acceleration/deceleration involved.
    Just a silly mistake that's all .




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    If no acceleration, use d=vt, if constant acceleration use suvat, if changing acceleration, go to university.
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    (Original post by mik1a)
    If no acceleration, use d=vt, if constant acceleration use suvat, if changing acceleration, go to university.
    lol!
 
 
 
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