At a certain time a particle had a speed of 46 m/s in the positive x direction, and 9.1 s later its speed was 67 m/s in the opposite directi

Question

At a certain time a particle had a speed of 46 m/s in the positive x direction, and 9.1 s later its speed was 67 m/s in the opposite direction. What was the average acceleration of the particle during this 9.1 s interval

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Khải Quang 6 months 2021-08-04T13:06:45+00:00 1 Answers 2 views 0

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    2021-08-04T13:08:05+00:00

    Answer:

    The average acceleration of the particle during this 9.1 s interval is  a=2.31 ms^{-2}.

    Explanation:

    The expression for the average acceleration of the particle is as follows;

    a_{vg}=\frac{v-u}{t'-t}

    Here, a_{vg} is the average acceleration, v is the final speed, u is the initial speed, t is the initial time and t’ is the final time.

    It is given in the problem that at a certain time a particle had a speed of 46 m/s in the positive x-direction, and 9.1 s later its speed was 67 m/s in the opposite direction.

    Put u=46 ms^{-1}, v=67 ms^{-1}, t= 0 and t’=9.1 s in the above expression.

    a_{vg}=\frac{67-46}{9.1-0}

    a=2.31 ms^{-2}

    Therefore, the average acceleration of the particle during this 9.1 s interval is  a=2.31 ms^{-2}.

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