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luda_lava [24]
3 years ago
14

A police car is located 50 feet to the side of a straight road. A red car is driving along the road in the direction of the poli

ce car and is 140 feet up the road from the location of the police car. The police radar reads that the distance between the police car and the red car is decreasing at a rate of 70 feet per second. How fast is the red car actually traveling along the road?

Physics
1 answer:
NNADVOKAT [17]3 years ago
6 0

Answer:74.33 feet per sec

Explanation:

Given

Police car is 50 feet side of road

Red car is 140 feet up the road

Distance between them is decreasing at the rate of 70 feet per sec

From figure

x^2+y^2=z^2

z=\sqrt{140^2+50^2}

z=148.66 feet

as the red car is moving

therefore its velocity magnitude is given by

\frac{\mathrm{d} x}{\mathrm{d} t}

2x\times \frac{\mathrm{d} x}{\mathrm{d} t}+0=2z\frac{\mathrm{d} z}{\mathrm{d} t}

x\frac{\mathrm{d} x}{\mathrm{d} t}=z\frac{\mathrm{d} z}{\mathrm{d} t}

140\times \frac{\mathrm{d} x}{\mathrm{d} t}=148.66\times 70

\frac{\mathrm{d} x}{\mathrm{d} t}=74.33 feet\ per\ sec

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Answer:

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Answer:

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\sum F=ma

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F_{g} =G\frac{Mm}{r^{2}}

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F_{c}=ma_{c}

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a_{c} = ( \frac{2\pi}{T})^{2}r

or:

a_{c}=\frac{4\pi^{2}r}{T^{2}}

so:

F_{c}=m(\frac{4\pi^{2}r}{T^{2}})

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\sum F=ma

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in this case we can get rid of the mass of the planet, so we get:

G\frac{M}{r^{2}}=(\frac{4\pi^{2}r}{T^{2}})

we can now solve this for T^{2} so we get:

T^{2} = \frac{4\pi ^{2}r^{3}}{GM}

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\frac{4\pi^{2}r^{3}}{GM_{1}}>\frac{4\pi^{2}r^{3}}{GM_{2}}

we can get rid of all the constants so we end up with:

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M_{2}>M_{1}

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