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8 August, 05:47

A minivan is tested for acceleration and braking. In the street-start acceleration test, the elapsed time is 8.6 s for a velocity increase from 10 km/h to 100 km/h. In the braking test, the distance traveled is 44 m during braking to a stop from 100 km/h. Assume constant values of acceleration and deceleration.

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  1. 8 August, 06:06
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    The question is incomplete. Here is the complete question:

    A minivan is tested for acceleration and braking. In the street-start acceleration test, the elapsed time is 8.6 s for a velocity increase from 10 km/h to 100 km/h. In the braking test, the distance traveled is 44 m during braking to a stop from 100 km/h. Assume constant values of acceleration and deceleration. Determine

    (a) the acceleration during the street-start test,

    (b) the deceleration during the braking test.

    Answer:

    (a) 37500 km/h²

    (b) 113636.36 km/h²

    Explanation:

    part (a)

    Because it is given that we can assume constant acceleration therefore we can use the following equation of motion:

    v = u + (a) (t)

    where v is final velocity, u is initial velocity, a is acceleration and t is time change

    Given in the question:

    v = 100km/h

    u = 10 km/h

    t = 8.6 sec (changing to hours)

    t = 0.0024 hours (round off to 4 decimal places)

    100 = 10 + (a x 0.0024)

    Rearranging the equation to find value of a

    a = (100 - 10) / 0.0024

    a = 37500 km/h² (Answer)

    part (b)

    Now we can use the following equation to find deceleration

    2 (a) (s) = v² - u²

    Where a is acceleration, s is distance travelled, v is final velocity and u is initial velocity

    Given in the question

    s = 44 m

    changing to km

    s = 0.044 km

    v = 0 km/h (because it stops)

    u = 100 km/h

    2 (a) (0.044) = (0) ² - (100) ²

    0.088 (a) = 0 - 10000

    a = - 10000/0.088

    a = - 113636.36 km/h2

    The negative sign in the answer shows that it is deceleration

    Therefore deceleration = 113636.36 km/h² (Answer)
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