Physics
Physics
5th Edition
ISBN: 9781260486919
Author: GIAMBATTISTA
Publisher: MCG
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Chapter 3, Problem 47P

(a)

To determine

The average velocity of the car in the whole trip.

(a)

Expert Solution
Check Mark

Answer to Problem 47P

The average velocity of the car in the whole trip is 3.33m/s, 45° north of east_.

Explanation of Solution

Given that the radius of the circular path is 20.0m, and the time taken for covering three quarters of the circle is 8.50s with constant speed.

The vector diagram indicating the initial velocity vi, and the final velocity vf is given in Figure 1. The center of the circular path is considered as the origin.

Physics, Chapter 3, Problem 47P

The +y direction represents the north, and +x direction represents the east.

Write the expression for the displacement vector of the car.

Δr=rfri (I)

Here, Δr is the displacement vector, rf is the final position vector, and ri is the initial position vector.

The initial position vector of the car is 20.0mx^, and the final position vector is 20.0my^.

Write the expression for the magnitude of the displacement vector.

|Δr|=rx2+ry2 (II)

Here, rx is the of x-component of Δr, and ry is the of y-component of Δr.

The average velocity will be in the same direction of the displacement vector.

Write the expression for the angle θ that the displacement vector (as well as the average velocity vector) makes with the +x-direction (east).

θ=tan1(ryrx) (III)

Write the expression for the magnitude of the average velocity.

|vav|=|Δr|Δt (IV)

Here, vav is the average velocity vector, and Δt is the time duration.

Conclusion:

Substitute 20.0mx^ for rf, and 20.0my^ for ri in equation (I) to find Δr.

Δr=20.0mx^(20.0my^)=20.0mx^+20.0my^

Substitute 20.0m for rx, and 20.0m for ry in equation (II) to find |Δr|.

|Δr|=(20.0m)2+(20.0m)2=28.3m

Substitute 20.0m for rx, and 20.0m for ry in equation (III) to find θ.

θ=tan1(20.0m20.0m)=+45°45°north of east

Substitute 28.3m for |Δr|, and 8.50s for Δt in equation (IV) to find |vav|.

|vav|=28.3m8.50s=3.33m/s

Combine the magnitude and direction of the average velocity vector.

vav=3.33m/s, 45° north of east.

Therefore, the average velocity of the car in the whole trip is 3.33m/s, 45° north of east_.

(b)

To determine

The average acceleration of the car in the whole trip.

(b)

Expert Solution
Check Mark

Answer to Problem 47P

The average acceleration of the car in the whole trip is 1.84m/s2, 45° south of east_.

Explanation of Solution

It is obtained that the average velocity of the car in the whole trip is 3.33m/s, 45° north of east.

Write the expression for the average acceleration.

aav=vfviΔt (V)

Here, aav is the average acceleration, vf is the final velocity, and vi is the initial velocity.

Since the car covers three quarters of the circle, the distance travelled is obtained as,

d=(3/4)2πr (VI)

Here, d is the distance covered, and r is the radius of the circular track.

The initial velocity is directed west and the final velocity is directed south. The expression for the initial and final velocity is obtained as,

vi=dΔtx^

vf=dΔty^

Use the expression for vi and vf in equation (V).

aav=(dΔty^)(dΔtx^)Δt=dΔt(x^y^)Δt=d(Δt)2(x^y^) (VII)

Use equation (VI) in (VII).

aav=(3/4)2πr(Δt)2(x^y^)=3πr2(Δt)2(x^y^) (VIII)

Write the expression for the magnitude of average acceleration.

|aav|=ax2+ay2 (IX)

Here, ax is the x-component of aav, and ay is the y-component of aav.

Conclusion:

From equation (VIII) it is clear that the average acceleration vector has x-component 3πr2(Δt)2, and y-component (3πr2(Δt)2).

Substitute (3πr/2(Δt)2) for ax, and (3πr/2(Δt)2) for ay in equation (IX) to obtain the expression for |aav|.

|aav|=(3πr2(Δt)2)2+(3πr2(Δt)2)2=3πr2(Δt)2 (X)

Write the expression for the angle θ that the average acceleration vector makes with the +x-direction (east).

θ=tan1(ayax) (XI)

Substitute 20.0m for r, and 8.50s for Δt in equation (X) to find |aav|.

|aav|=3π(20.0m)2(8.50s)2=1.84m/s2

Substitute 1.84m/s2 for ax,and 1.84m/s2 for ay in equation (XI) to find θ

θ=tan1(1.84m/s21.84m/s2)=45°45°south of east

Combine the magnitude and direction of the average acceleration vector.

aav=1.84m/s2, 45° south of east.

Therefore, the average acceleration of the car in the whole trip is 1.84m/s2, 45° south of east_.

(c)

To determine

The reason for which a car moving at constant speed in a circular path has a nonzero average acceleration.

(c)

Expert Solution
Check Mark

Answer to Problem 47P

The change in direction of velocity requires an acceleration, which leads to the presence of a nonzero acceleration to a car moving in a circular path with constant speed.

Explanation of Solution

The direction of instantaneous velocity of the car moving in a circular path, changes in each instant. The direction of velocity at a point on the circular path will be along the tangent drawn at each of those points. Even though the magnitude of velocity (which is the speed) of motion is constant, due to the direction change of the velocity vector, the car gains an acceleration.

Therefore, the change in direction of velocity requires an acceleration, which leads to the presence of a nonzero acceleration to a car moving in a circular path with constant speed.

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Chapter 3 Solutions

Physics

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