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21 April, 18:32

A race car starts from rest and travels east along a straight and level track. For the first 5.0 s of the car's motion, the eastward component of the car's velocity is given by vx (t) = (0.890m/s3) t2.

Part A

What is the acceleration of the car whenvx = 14.1 m/s?

Express your answer with the appropriate units.

ax=

+2
Answers (1)
  1. 21 April, 20:31
    0
    a_x (3.98) = 7.0844 m/s^2

    Explanation:

    Given:

    - The eastward component of car's velocity is given as follows:

    v_x (t) = 0.890*t^2

    For, 0 s < t < 5 s

    Find:

    What is the acceleration of the car when v_x = 14.1 m/s?

    Solution:

    - We will first compute the time t at which the velocity v_x component equals 14.1 m / s:

    14.1 = 0.890*t^2

    t = sqrt (15.8427)

    t = 3.98 s

    - Now use the relation of v_x and derivative to obtain a_x acceleration in eastward direction:

    a_x (t) = dv_x / dt

    a_x (t) = 1.78*t

    - Evaluate at t = 3.98 s:

    a_x (3.98) = 1.783.98

    a_x (3.98) = 7.0844 m/s^2
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