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sweet [91]
2 years ago
11

If the coefficient of kinetic friction between tires and dry pavement is 0.98, what is the shortest distance in which you can st

op an automobile by locking the brakes when traveling at 34.7 m/s
Physics
1 answer:
Ilia_Sergeevich [38]2 years ago
7 0

Answer:

The shortest distance is 62.7 m

Explanation:

Given;

coefficient of kinetic friction, μk = 0.98

initial velocity, u = 34.7 m/s

Frictional force on the tire;

Fk = -μkN

where;

N is normal reaction = mg

ma = -μkN

ma = -μkmg

a = -μkg

a = - 0.98 x 9.8 = -9.604 m/s²

The shortest distance in which you can stop an automobile by locking the brakes:

Apply equation of motion;

v² = u² + 2ax

where;

v is the final velocity

u is the initial velocity

a is the acceleration of the  automobile

0 = 34.7² + 2(-9.604)x

0 = 1204.09 - 19.208x

19.208x = 1204.09

x = 1204.09/19.208

x = 62.7 m

Therefore, the shortest distance in which you can stop an automobile by locking the brakes when traveling at 34.7 m/s is 62.7 m

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Answer:

<em> The planes average acceleration in magnitude and direction = 8.846 m/s² moving east</em>

Explanation:

Acceleration: This can be defined as the rate of change of velocity. The S.I Unit of acceleration is m/s². Acceleration is a vector quantity because it can be represented both in magnitude and in direction.

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a = v/t.................................... Equation 1

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<em>Given: v = 115 m/s, t = 13.0 s</em>

<em>Substituting these values into equation 1</em>

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<em>Thus the planes average acceleration in magnitude and direction = 8.846 m/s² moving east</em>

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Refer to the diagram shown below.

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