A particle is moving with a velocity of 60.0 m/s in the positive x direction at t= 0. Between t= 0 and t= 15.0 s the velocity decreases unif

Question

A particle is moving with a velocity of 60.0 m/s in the positive x direction at t= 0. Between t= 0 and t= 15.0 s the velocity decreases uniformly to zero. What was the acceleration during this 15.0-s interval? What is the significance of the sign of your answer?

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Thạch Thảo 4 years 2021-08-24T03:03:22+00:00 1 Answers 51 views 0

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  1. Verity
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    2021-08-24T03:04:45+00:00
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    Answer:

    Acceleration of the object is given as

    a = - 4 m/s^2

    so here negative sign shows that the direction of acceleration is opposite to the velocity of the object

    Explanation:

    As we know that velocity decreases uniformly from maximum to zero

    so here in this uniform deceleration we know that

    v_i = 60 m/s

    v_f = 0

    t = 15 s

    now by using the equation of kinematics we have

    v_f - v_i = at

    0 - 60 = a(15)

    a = - 4 m/s^2

    so here negative sign shows that the direction of acceleration is opposite to the velocity of the object

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