Traveling at a speed of 16.0 m/s, the driver of an automobile suddenly locks the wheels by slamming on the brakes. The coefficient of kineti

Question

Traveling at a speed of 16.0 m/s, the driver of an automobile suddenly locks the wheels by slamming on the brakes. The coefficient of kinetic friction between the tires and the road is 0.600. What is the speed of the automobile after 1.25 s have elapsed

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Thu Cúc 2 months 2021-07-23T06:06:55+00:00 1 Answers 2 views 0

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    2021-07-23T06:08:44+00:00

    Answer:

    The speed of the automobile after 1.25 s have elapsed is 8.65 m/s

    Explanation:

    Given;

    initial velocity of the driver, v_i = 16 m/s

    coefficient of kinetic friction, \mu_k = 0.6

    time of motion, t = 1.25s

    The final velocity of the driver is calculated as follows;

    a = \frac{\Delta v}{\Delta t} = - \frac{v_f - v_i}{\Delta t}\\\\ (the \ negative \ sign \ indicates \ that \ the \ final \ velocity \ will \,be\  smaller \\\ than  \ the \ initial \ velocity \ because \ of \ limiting \ frictiona\  force)\\\\v_f- v_i = -a\Delta t\\\\v_f = v_i - at\\\\Recall \ from \ Newton's \ law; \ F_k = \mu_k mg = ma \\ \ a = \mu_kg\\\\v_f = v_i - (\mu_k g )t\\\\v_f = 16 - (0.6 \times 9.8) 1.25\\\\v_f = 16 - 7.35\\\\v_f = 8.65 \ m/s

    Therefore, the speed of the automobile after 1.25 s have elapsed is 8.65 m/s

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