A car accelerates uniformly from rest and after 12 seconds has covered 40m. What are its acceleration and its final velocity?

Question

A car accelerates uniformly from rest and after 12 seconds has covered 40m.
What are its acceleration and its final velocity?

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Kim Chi 3 years 2021-07-26T13:19:25+00:00 1 Answers 160 views 0

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    2021-07-26T13:20:48+00:00

    Answer:

    a\approx 0.56 \,\frac{m}{s^2}

    v_f=6.72\,\frac{m}{s}

    Explanation:

    This is a motion under constant acceleration, so we can use the kinematic equation for the distance covered:

    x_f-x_i=v_i\,t+\frac{1}{2} a\,t^2\\40=0+\frac{1}{2} a\,(12)^2\\40=72 \,a\\a=\frac{40}{72} \,\frac{m}{s^2} \\a\approx 0.56 \,\frac{m}{s^2}

    Now, the final velocity can be calculated via:

    v_f=v_i+a\,t\\v_f=0+0.56\,(12)\\v_f=6.72\,\frac{m}{s}

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