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3 years ago

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A car initially traveling at 32.4 m/s undergoes a constant negative acceleration of magnitude 1.70 m/s2 after its brakes are app

lied. (a) How many revolutions does each tire make before the car comes to a stop, assuming the car does not skid and the tires have radii of 0.330 m?

1 answer:

8090 [49]3 years ago

8 0

<h2>The car tire makes 149 revolutions before coming to stop.</h2>

Explanation:

a) We have equation of motion v² = u² + 2as

Initial velocity, u = 32.4 m/s

Acceleration, a = -1.70 m/s²

Final velocity, v = 0 m/s

Substituting

v² = u² + 2as

0² = 32.4² + 2 x -1.70 x s

s = 308.75 m

Distance traveled before stopping is 308.75 m

Radius of tire = 0.330 m

Circumference of tire = 2πx 0.33 = 2.07 m

1 revolution = 2.07 m

$\texttt{Number of revolutions = }\frac{308.75}{2.07}=148.91$

The car tire makes 149 revolutions before coming to stop.

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Answer:

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Where:

$v_{o}$ - Initial velocity, measured in meters per second.

$y$ - Final height, measured in meters.

$y_{o}$ - Initial height, measured in meters.

$t$ - Time, measured in seconds.

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(Positive sign - Package is moving upward, Negative sign - Package is moving downward)

The initial height is now cleared:

$y_{o} = y - v_{o}\cdot t - \frac{1}{2}\cdot g \cdot t^{2}$

Given that $y = 0\,m$ , $v_{o} = 15\,\frac{m}{s}$ , $g = -9.807\,\frac{m}{s^{2}}$ and $t = 16\,s$ , the final height of the package is:

$y_{o} = 0\,m - \left(15\,\frac{m}{s} \right)\cdot (16\,s) - \frac{1}{2}\cdot \left(-9.807\,\frac{m}{s^{2}} \right) \cdot (16\,s)^{2}$

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