Physics
Physics
3rd Edition
ISBN: 9780073512150
Author: Alan Giambattista, Betty Richardson, Robert C. Richardson Dr.
Publisher: McGraw-Hill Education
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Chapter 5, Problem 49P

(a)

To determine

The speed of the car at 1/4 of the circumference of the car traveled.

(a)

Expert Solution
Check Mark

Answer to Problem 49P

The speed of the car is 17.7m/s.

Explanation of Solution

Since the car moves with constant tangential acceleration with the circumference C=2πr.

Write the equation for the distance traveled by the car.

Δx=C4 (I)

Here, Δx is the distance traveled by the car and C is the circumference of the car traveled.

Replace 2πr for C in the equation (I).

Δx=2πr4=πr2 (II)

Here, the radius of the circular path of the car traveled is r.

Write the equation of motion for the speed of the car.

vt2=u2+2atΔx (III)

Here, vt is the final tangential speed of the car, u is the initial speed of the car, and at is the tangential acceleration of the car traveled.

Conclusion:

Substitute equation (II) in equation (III) and 0m/s for u.

vt2=(0m/s)2+2at(πr2)=atπrvt=atπr

Substitute 2.00m/s2 for at and 50.0m for r in above equation.

vt=(2.00m/s2)(3.14)(50.0m)=17.7m/s

Therefore, the speed of the car is 17.7m/s.

(b)

To determine

The radial acceleration of the car at 1/4 of the circumference of car traveled.

(b)

Expert Solution
Check Mark

Answer to Problem 49P

The radial acceleration of the car is 6.28m/s2.

Explanation of Solution

Write the equation for radial acceleration of the car.

ar=vt2r                                                                                     (IV)

Here, ar is the radial acceleration of the car traveled at the circumference of 1/4.

Conclusion:

Substitute aπr for vt2 in equation (IV).

ar=atπrr=atπ (V)

Substitute 2.00m/s2 for at in equation (V).

ar=(2.00m/s2)(3.14)=6.28m/s2

Therefore, the radial acceleration of the car is 6.28m/s2.

(c)

To determine

The total acceleration of the car.

(c)

Expert Solution
Check Mark

Answer to Problem 49P

The total acceleration of the car is 6.59m/s2 directed by an angle 17.7°.

Explanation of Solution

The figure 1 shows the total acceleration exerted on the car.

Physics, Chapter 5, Problem 49P

Write the equation for total acceleration of the car.

a=ar2+at2                                                                                    (VI)

Here, a is the total acceleration of the car traveled at the circumference of 1/4.

Write the equation for the direction of the car travelled.

tanθ=ayax (VII)

Conclusion:

Substitute atπ for ar in equation (VI).

a=at2π2+at2=at2(π2+1)=at(π2+1) (VIII)

Substitute 2.00m/s2 for at in equation (VIII).

a=(2.00m/s2)(3.142+1)=6.59m/s2

Therefore, the total acceleration of the car is 6.59m/s2.

Let us calculate the direction of the car travelled at the point of circumference 1/4.

Consider +y can be taken as north direction and +x can be taken as east.

Since the acceleration for +x is ax=at, then acceleration for +y is ay=ar.

Rewrite the equation (VII) for θ.

θ=tan1ayax=tan1(arat)=tan1(πatat)=tan1(π)

Substitute 3.14 for π in above equation.

θ=tan1(3.14)=72.3°ϕ=(90°72.3°)=17.7°

Therefore, the total acceleration of the car is 6.59m/s2 directed along the east of south direction at an angle 17.7°.

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Chapter 5 Solutions

Physics

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