A 2500 kg car traveling to the north is slowed down uniformly from an initial velocity of 20 m/s by a 5620 N braking force acting opposite t

Question

A 2500 kg car traveling to the north is slowed down uniformly from an initial velocity of 20 m/s by a 5620 N braking force acting opposite the car’s motion. How far does the car move during 2.5 s? *

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Eirian 5 years 2021-08-10T14:06:08+00:00 1 Answers 155 views 0

Answers ( )

  1. kimcuc
    0
    2021-08-10T14:07:34+00:00
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    Answer:

    the distance traveled by the car is 42.98 m.

    Explanation:

    Given;

    mass of the car, m = 2500 kg

    initial velocity of the car, u = 20 m/s

    the braking force applied to the car, f = 5620 N

    time of motion of the car, t = 2.5 s

    The decelaration of the car is calculated as follows;

    -F = ma

    a = -F/m

    a = -5620 / 2500

    a = -2.248 m/s²

    The distance traveled by the car is calculated as follows;

    s = ut + ¹/₂at²

    s = (20 x 2.5) + 0.5(-2.248)(2.5²)

    s = 50 – 7.025

    s = 42.98 m

    Therefore, the distance traveled by the car is 42.98 m.

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