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The speed of the particle

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The acceleration of the particle

a ( t ) = e −5 t ( sin t 5 cos t ) 5 e −5 t ( cos t 5 sin t ) , e −5 t ( cos t 5 sin t ) + 5 e −5 t ( sin t + 5 cos t ) , 100 e −5 t

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Find the maximum speed of a point on the circumference of an automobile tire of radius 1 ft when the automobile is traveling at 55 mph.

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A projectile is shot in the air from ground level with an initial velocity of 500 m/sec at an angle of 60° with the horizontal. The graph is shown here:

This figure is a curve in the fourth quadrant. The curve is decreasing. It begins at the origin and decreases into the fourth quadrant. Answer the following questions.

At what time does the projectile reach maximum height?

44.185 sec

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What is the approximate maximum height of the projectile?

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At what time is the maximum range of the projectile attained?

t = 88.37 sec

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What is the maximum range?

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What is the total flight time of the projectile?

88.37 sec

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A projectile is fired at a height of 1.5 m above the ground with an initial velocity of 100 m/sec and at an angle of 30° above the horizontal. Use this information to answer the following questions:

Determine the maximum height of the projectile.

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Determine the range of the projectile.

The range is approximately 886.29 m.

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A golf ball is hit in a horizontal direction off the top edge of a building that is 100 ft tall. How fast must the ball be launched to land 450 ft away?

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A projectile is fired from ground level at an angle of 8° with the horizontal. The projectile is to have a range of 50 m. Find the minimum velocity necessary to achieve this range.

v = 42.16 m/sec

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Prove that an object moving in a straight line at a constant speed has an acceleration of zero.

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The acceleration of an object is given by a ( t ) = t j + t k . The velocity at t = 1 sec is v ( 1 ) = 5 j and the position of the object at t = 1 sec is r ( 1 ) = 0 i + 0 j + 0 k . Find the object’s position at any time.

r ( t ) = 0 i + ( 1 6 t 3 + 4.5 t 14 3 ) j + ( t 3 6 1 2 t + 1 3 ) k

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Find r ( t ) given that a ( t ) = −32 j , v ( 0 ) = 600 3 i + 600 j , and r ( 0 ) = 0 .

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Find the tangential and normal components of acceleration for r ( t ) = a cos ( ω t ) i + b sin ( ω t ) j at t = 0 .

a T = 0 , a N = a ω 2

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Given r ( t ) = t 2 i + 2 t j and t = 1 , find the tangential and normal components of acceleration.

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For each of the following problems, find the tangential and normal components of acceleration.

r ( t ) = e t cos t , e t sin t , e t . The graph is shown here:

This figure is a curve in 3 dimensions. It is inside of a box. The box represents an octant. The curve begins in the bottom of the box, from the lower left, and bends through the box to the other side, in the upper left.

a T = 3 e t , a N = 2 e t

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r ( t ) = cos ( 2 t ) , sin ( 2 t ) , 1

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r ( t ) = 2 t , t 2 , t 3 3

a T = 2 t , a N = 4 + 2 t 2

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r ( t ) = 2 3 ( 1 + t ) 3 / 2 , 2 3 ( 1 t ) 3 / 2 , 2 t

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r ( t ) = 6 t , 3 t 2 , 2 t 3

a T 6 t + 12 t 3 1 + t 4 + t 2 , a N = 6 1 + 4 t 2 + t 4 1 + t 2 + t 4

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r ( t ) = t 2 i + t 2 j + t 3 k

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r ( t ) = 3 cos ( 2 π t ) i + 3 sin ( 2 π t ) j

a T = 0 , a N = 2 3 π

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Find the position vector-valued function r ( t ) , given that a ( t ) = i + e t j , v ( 0 ) = 2 j , and r ( 0 ) = 2 i .

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The force on a particle is given by f ( t ) = ( cos t ) i + ( sin t ) j . The particle is located at point ( c , 0 ) at t = 0 . The initial velocity of the particle is given by v ( 0 ) = v 0 j . Find the path of the particle of mass m . (Recall, F = m · a . )

r ( t ) = ( −1 m cos t + c + 1 m ) i + ( sin t m + ( v 0 + 1 m ) t ) j

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An automobile that weighs 2700 lb makes a turn on a flat road while traveling at 56 ft/sec. If the radius of the turn is 70 ft, what is the required frictional force to keep the car from skidding?

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Using Kepler’s laws, it can be shown that v 0 = 2 G M r 0 is the minimum speed needed when θ = 0 so that an object will escape from the pull of a central force resulting from mass M . Use this result to find the minimum speed when θ = 0 for a space capsule to escape from the gravitational pull of Earth if the probe is at an altitude of 300 km above Earth’s surface.

10.94 km/sec

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Practice Key Terms 6

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Source:  OpenStax, Calculus volume 3. OpenStax CNX. Feb 05, 2016 Download for free at http://legacy.cnx.org/content/col11966/1.2
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