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Hi can someone please help me with this question, and step by step if possible? Thanks?

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a car traveling 45 km/h slows down at a constant 0.50 m/s^2 just by "letting up on the gas". Calculate the distance the car coasts before it stops, the time it takes to stop, and the distnace it travels during the first and fifth seconds.

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  1. u = 45 kmph = 12.5 m/s

    a = - 0.5 m/s^2

    v = 0 m/s

    s = (v^2 - u^2)/2a = 156.25 m

    t = (v-u)/a = 25 s

    Distance traveled in the nth second is given by the difference of the distance traveled in the first n seconds and the distance traveled in the first (n-1) seconds. Work out the expression yourself.

    I think it is as follows :

    Sn = u + a/2*(2n - 1)

    Put n = 1 and 5:

    S1 = 12.25 m

    S5 = 10.25 m

    Hope this helps.

    your_guide123@yahoo.com

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