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Problem 11.190

The driver of an automobile decreases her speed at aconstant rate from 45 to 30 mi/h over a distance of 750 ftalong a curve of 1500-ft radius. Determine the magnitudeof the total acceleration of the automobile after the

automobile has traveled 500 ft along the curve.

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Solving Problems on Your OwnProblem 11.190

The driver of an automobile decreases her speed at aconstant rate from 45 to 30 mi/h over a distance of 750 ftalong a curve of 1500-ft radius. Determine the magnitude

of the total acceleration of the automobile after theautomobile has traveled 500 ft along the curve.

1. Use tangential and normal components: These components

are used when the particle travels along a circular path. Theunit vector e t is tangent to the path (and thus aligned with thevelocity) while the unit vector e n is directed along the normal tothe path and always points toward its center of curvature.

2. Determine the tangential acceleration: For a constanttangential acceleration, the acceleration can be determined from

v dv = a t dxv o

v

x o

x

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Solving Problems on Your OwnProblem 11.190

The driver of an automobile decreases her speed at aconstant rate from 45 to 30 mi/h over a distance of 750 ftalong a curve of 1500-ft radius. Determine the magnitude

of the total acceleration of the automobile after theautomobile has traveled 500 ft along the curve.

3. Determine the normal acceleration: For known velocity andradius of curvature, the tangential acceleration is determine by

a n =

4. Determine the magnitude of the total acceleration: Themagnitude of the total acceleration is given by

a = a t 2 + a n 2

v 2

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Problem 11.190 Solution

1500 ft

45 mi/h

30 mi/h

a t 7 5

0 f t

Determine the tangential acceleration.

45 mi/h = 44 ft/s45 mi/h = 66 ft/s

vo

v

xo

xv dv = a t dx

66

44

0

750v dv = a t dx

( 44 2 - 66 2 ) = a t ( 750 - 0 )21

a t = - 1.613 ft/s 2

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a t = -1.613 ft/s2

1500 ft

45 mi/h

v1 a n

5 0 0 f t

Problem 11.190 Solution

The speed after the automobiletraveled 500 ft.

vo

v

xo

xv dv = a t dx

66

v1

0

500v dv = -1.613 dx

21 (v 12 - 66 2 ) = -1.613 ( 500 - 0 )

v1 = 52.4 ft/s

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a t = -1.613 ft/s2

1500 ft

45 mi/h

52.4 ft/s a n

5 0 0 f t

Problem 11.190 Solution

Determine the normal acceleration.

a n =v 2

a n = = 1.828 ft/s 2( 52.4 ft/s )2

1500 ft

Determine the magnitude of the total acceleration.

a = a t 2 + a n 2

a = ( - 1.613 ft/s 2 )2 + ( 1.828 ft/s 2 )2

a = 2.44 ft/s 2