Physics For Scientists And Engineers With Modern Physics, 9th Edition, The Ohio State University
Physics For Scientists And Engineers With Modern Physics, 9th Edition, The Ohio State University
9th Edition
ISBN: 9781305372337
Author: Raymond A. Serway | John W. Jewett
Publisher: Cengage Learning
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Chapter 4, Problem 1P

(a)

To determine

The total vector displacement.

(a)

Expert Solution
Check Mark

Answer to Problem 1P

The displacement is 4.87 km_ at 28.6° South of West.

Explanation of Solution

Draw the figure according to given condition as.

Physics For Scientists And Engineers With Modern Physics, 9th Edition, The Ohio State University, Chapter 4, Problem 1P

Consider that system has three vectors.

Write the expression for first trip from point A to B.

    d1=v1t1                                                                                                           (I)

Here, d1 is distance for first trip, v1 is speed for first trip and t1 is time for first trip.

Write the expression for second trip from point B to C .

  d2=v2t2                                                                                                       (II)

Here, d2 is distance for second trip, v2 is speed for second trip and t2 is time for second trip.

Write the expression for third trip from point C to D.

    d3=v3t3                                                                                                       (III)

Here, d3 is distance for third trip, v3 is speed for third trip and t3 is time for third trip.

Write the expression for total displacement in three trips.

    d=d1+d2+d3                                                                                           (IV)

Here, d is total displacement vector.

Write the expression for magnitude of the displacement.

    |d|=(dx2)+(dy2)                                                                                      (V)

Here, dx is the x-component of distance and dy is the y-component of distance.

Write the expression for direction of vector d as.

    θ=tan1(dydx)                                                                                            (VI)

Here, θ is angle between two components.

Conclusion:

Substitute 20.0j^ m/s for v1 and 3.00 min for t1 in equation (II)

    d1=(20.0j^ m/s(3.00 min))=(20.0 m/s(3.00 min(60 s1 min)))j^=3600j^ m

Substitute 25.0i^ m/s for v2 and 2.00 min for t2 in equation (III)

  d2=(25.0i^ m/s(2.00 min))=(25.0 m/s(2.00 min(60 s1 min)))i^=3000i^ m

Substitute (30.0cos45°i^+30sin45°j^) m/s for v3 and 1.00 min for t3 in equation (IV)

  d3=[(30.0cos45°i^+30sin45°j^) m/s](1.00 min)=[(30.0cos45° m/s(1.00 min(60 s1 min)))i^+(30.0sin45° m/s(1.00 min(60 s1 min)))]j^=(1272.8i^ +1272.8j^)m

Substitute 3600j^ m for d1, 3000i^ m for d2 and (1272.8i^ +1272.8j^)m for d3 in equation (V).

    d=(3600j^ m3000i^ m)+(1272.8i^ m+1272.8j^ m)=4272.8i^ m2327.2j^ m

Substitute 4272.8i^ m for dx and 2327.2j^ m for dy in equation (VI)

    d=(4272.8i^)2+(2327.2j^)2=4865.45 m(1km1000 m)4.87 km

Substitute 4272.8i^ m for dx and 2327.2j^ m for dy in equation (VII).

    θ=tan1(2327.2m4272.8m)=tan1(0.54465)=28.57°28.6°

Thus the displacement is 4.87 km_ at 28.6° South of West.

(b)

To determine

The average speed.

(b)

Expert Solution
Check Mark

Answer to Problem 1P

The average speed is 23.3 m/s_.

Explanation of Solution

Write the expression for average speed as.

    vavg=d1+d2+d3t1+t2+t3                                                                                 (VIII)

Here, vavg is the average speed.

Conclusion:

Substitute 3600 m for d1, 3000 m for d2, 1800 m for d3, 3.00 min for t1, 2.00 min for t2 and 1.00 min for t3 in equation (VIII)

    vavg=(3600 m+3000 m+1800 m)(3.00 min+2.00 min+1.00 min)=(8400 m)(6.00 min)=(8400 m)(6.00 min(60 s1 min))=8400 m360 s

Simplify further above equation as.

    vavg=23.3 m/s

Thus, the average speed is 23.3 m/s_.

(c)

To determine

The average velocity.

(c)

Expert Solution
Check Mark

Answer to Problem 1P

The average velocity is 13.5 m/s_ at 28.6° South of West.

Explanation of Solution

Write the expression for average velocity as.

    vavg=dt1+t2+t3                                                                                           (IX)

Here, vavg is the average velocity.

Conclusion:

Substitute 4865.45 m for d, t1, 2.00 min for t2 and 1.00 min for t3 in equation (IX)

    vavg=(4865.45 m)(3.00 min+2.00 min+1.00 min)=(4865.45 m)(6.00 min)=(4865.45 m)(6.00 min(60 s1 min))=(4865.45 m)(360 s)

Simplify above equation as.

    vavg=13.5 m/s

Thus, the average velocity is 13.5 m/s_ at 28.6° South of West.

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Chapter 4 Solutions

Physics For Scientists And Engineers With Modern Physics, 9th Edition, The Ohio State University

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