A girl rolls a basketball across a basketball court. The ball slowly decelerates at a rate of -0.20m/s^2. If the initial velocity was 2.0m/s

Question

A girl rolls a basketball across a basketball court. The ball slowly decelerates at a rate of -0.20m/s^2. If the initial velocity was 2.0m/s and the ball rolled to a stop at 5.0sec after. 12:00, at what time did she start the ball rolling?​

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Amity 6 months 2021-07-19T00:53:18+00:00 1 Answers 63 views 0

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    2021-07-19T00:55:06+00:00

    By equation of motion :

    2as=v^2-u^2\\\\s=\dfrac{v^2-u^2}{2a}\\\\s=\dfrac{0^2-2^2}{2\times( -0.2)}\\\\s=10\ m

    Now, it is given that it stops after 5 seconds of motion and time at that point is 12:00.

    So, time when it started is 11 : 59 : 55.

    Therefore, distance travelled is 10 m.

    Hence, this is the required solution.

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Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )