ou have been called to testify as an expert witness in a trial involving a head-on collision. Car A weighs 1515 lb and was traveling eastwar

Question

ou have been called to testify as an expert witness in a trial involving a head-on collision. Car A weighs 1515 lb and was traveling eastward. Car B weighs 1125 lb and was traveling westward at 44.0 mph. The cars locked bumpers and slid eastward with their wheels locked for 17.5 ft before stopping. You have measured the coefficient of kinetic friction between the tires and the pavement to be 0.750 . How fast (in miles per hour) was car A traveling just before the collision

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Thiên Ân 4 years 2021-08-12T10:08:59+00:00 1 Answers 342 views 0

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  1. kiet
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    2021-08-12T10:10:12+00:00
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    Answer:

    67.22 m/s

    Explanation:

    First, find deceleration which is equivalent to, a=\mu_k g where \mu_k is coefficient of kinetic friction and g is acceleration due to gravity. Taking g as 32.2 then a=0.75*32.2=24.15 ft/s2

    From kinematics

    v^{2}=u^{2}-2as where v is final velocity and u is initial velocity. In this case, initial velocity is the common velocity and the final velocity is zero since they come to rest. S is the distance moved and a is deceleration. Substituting v with 0 and making u the subject then u=\sqrt{2as} and by substitution u=\sqrt{2\times 24.15\times 17.5}\approx 29.073 ft/s

    Converting ft per s to miles per hour, we multiply the above by 0.681818 hence u=29.073*0.681818=19.8225 mph

    From the law of conservation of linear momentum, p=mv where m is mass and v is velocity, the sum of initial and final momentum are equal. In this case

    m_1v_1+m_2v_2=(m_1+m_2)u

    By substitution considering the u is already known as 19.8225 mph

    1515v_1-(1125*44)=(1515+1125)*19.8225\\v_1=\frac {(1515+1125)*19.8225+(1125*44)}{1515}\approx 67.22\ mph

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