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3 years ago

What is the average velocity of a car address for my .7 m to 2 m away from the garage in 3 seconds

Physics

1 answer:

stepan [7]3 years ago

5 0

Answer:

$v=\frac{5}{3}$

Explanation:

So average velocity is calculated as

$average velocity=\frac{total displacement}{total time}$

Hence in our given case total displacement would be 5 m as the car moves a distance 7m to 2m.

Time taken for this moment is given to be 3 seconds.

Hence substituting back we get the average velocity as

$v=\frac{5}{3}$

Hence this is the average velocity in this case.

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Answer:

The answer to the question is;

The rate at which the distance between the two cars is changing is equal to 14.4 ft/sec.

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The distance between the two cars is given by the length of the hypotenuse side of a right angled triangle with the north being the y coordinate and the west being the x coordinate.

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The rate of change of the distance from their location 4 seconds after they commenced their journeys is given by;

Since s² = x² + y² we have

$\frac{ds^{2} }{dt} = \frac{dx^{2} }{dt} + \frac{dy^{2} }{dt}$

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$s\frac{ds }{dt} = x\frac{dx}{dt} + y\frac{dy }{dt}$

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Car B moving north = 12 ft/sec, which is the y direction and

Car A moving west = 8 ft/sec which is the x direction.

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$\frac{dy }{dt}$ = 12 ft/sec and

$\frac{dx}{dt}$ = 8 ft/sec

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$32.202 ft.\frac{ds }{dt} = 19 ft \times 8 \frac{ft}{sec} + 26ft\times 12\frac{ft}{sec}$ = 464 ft²/sec

$\frac{ds }{dt} = \frac{464\frac{ft^{2} }{sec} }{32.202 ft.}$ = 14.409 ft/sec ≈ 14.4 ft/sec to one place of decimal.

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If you notice any mistake in my english, please let me know, because I am not native.

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