Suppose a police officer is 1/2 mile south of an intersection, driving north towards the intersection at 40 mph. At the same time, another c

Question

Suppose a police officer is 1/2 mile south of an intersection, driving north towards the intersection at 40 mph. At the same time, another car is 1/2 mile east of the intersection, driving east (away from the intersection) at an unknown speed. The officer’s radar gun indicates 25 mph when pointed at the other car (that is, the straight-line distance between the officer and the other car is increasing at a rate of 25 mph). What is the speed of the other car?

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Ngọc Khuê 6 months 2021-07-17T21:11:13+00:00 1 Answers 46 views 0

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    2021-07-17T21:12:14+00:00

    Answer:

    75.36 mph

    Explanation:

    The distance between the other car and the intersection is,

    x=x_{0}+V t \\ x=\frac{1}{2}+V t

    The distance between the police car and the intersection is,

    y=y_{0}+V t

    y=\frac{1}{2}-40 t

    (Negative sign indicates that he is moving towards the intersection)

    Therefore the distance between them is given by,

    z^{2}=x^{2}+y^{2}(\text { Using Phythogorous theorem })

    z^{2}=\left(\frac{1}{2}+V t\right)^{2}+\left(\frac{1}{2}-40 t\right)^{2} \ldots \ldots \ldots(1)

    The rate of change is,

    2 z \frac{d z}{d t}=2\left(\frac{1}{2}+V t\right) V+2\left(\frac{1}{2}-40 t\right)(-40)

    2 z \frac{d z}{d t}=V+2 V^{2} t-40+3200 t \ldots \ldots \ldots

    Now finding z when t=0, from (1) we have

    z^{2}=\left(\frac{1}{2}+V(0)\right)^{2}+\left(\frac{1}{2}-40(0)\right)^{2}

    z^{2}=\frac{1}{4}+\frac{1}{4}=\frac{1}{2} \\ z=\sqrt{\frac{1}{2}} \approx 0.7071

    The officer’s radar gun indicates 25 mph pointed at the other car then, \frac{d z}{d t}=25 when t=0, from

    From (2) we get

    2(0.7071)(25)=V+2 V^{2}(0)-40+3200(0)

    2(0.7071)(25)=V+2 V^{2}(0)-40

    35.36=V-40

    V=35.36+40=75.36

    Hence the speed of the car is 75.36 mph

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