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17 August, 07:14

Consider two cars, a 700kg Porsche and a 600kg Honda Civic. The Porsche is speeding along at 40 m/s (mph) and the Civic is going half the speed at 20 m/s. If the two cars brake to a stop with the same constant acceleration, lets look at whether the amount of time required to come to a stop or the distance traveled prior to stopping is influenced by their initial velocity.

1. A car traveling 5m/s slams on its brakes, creating an acceleration of - 2 m/s^2. How far did the car travel after it applied its brakes?

2. The same car traveling for 10m/s applies the same acceleration of - 2 m/s^2. How far did the car travel after it applied its brakes?

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  1. 17 August, 07:31
    0
    To find the distance covered by the car after it applied brakes, we use 3rd equation of motion.

    2as = Vf² - Vi²

    s = (Vf² - Vi²) / 2a

    1.

    We have:

    Vi = Initial Velocity = 5 m/s

    Vf = Final Velocity = 0 m/s (Since, car finally stops)

    a = deceleration = - 2 m/s²

    s = distance covered by the car = ?

    Therefore,

    s = [ (0 m/s) ² - (5 m/s) ²]/2 ( - 2 m/s²)

    s = 6.25 m

    2.

    We have:

    Vi = Initial Velocity = 10 m/s

    Vf = Final Velocity = 0 m/s (Since, car finally stops)

    a = deceleration = - 2 m/s²

    s = distance covered by the car = ?

    Therefore,

    s = [ (0 m/s) ² - (10 m/s) ²]/2 ( - 2 m/s²)

    s = 25 m

    Hence, the distance traveled by the car is affected by the initial speed in accordance with a direct relationship.
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