Calc

The maker of an automobile advertises that it takes 13 seconds to accelerate from 25 kilometers per hour to 90 kilometers per hour. Assuming constant acceleration, compute the following.

The acceleration in m/s^2

The distance the car travels during the 13 seconds

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asked by Kelly
  1. if we have acceleration of a, then
    v = at + c
    case1: when t=0, v=25
    25 = 0+c ---> v = at + 25
    so when t = 13, v = 90
    90 = 13a + 25
    a = 65/13 = 5 m/s^2 -----> v = 5t + 25

    distance = (5/2)t^2 + 25t + k
    when t=0 , distance = 0 , so k = 0

    distance = (5/2)t^2 + 25t
    when t = 13
    distance = (5/2)(169) + 25(13) = 747.5 m

    also see the first four of the Related Questions below

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    posted by Reiny
  2. a = (V2-V1)/(t2-t1)

    V2 = 90,000/3600 = 25 m/s
    V1 = 25,000/3600 = 6.94

    t2-t1 = 13

    so a = 1.39 m/s^2
    ======================
    d = V1 t + (1/2)a t^2
    = 6.94 *13 + .695*169
    = 208 m

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    posted by Damon
  3. watch out, km/hour given

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    posted by Damon
  4. forgot to change the units from km/h to m/s

    go with Damon's answers

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    posted by Reiny

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