COLLEGE PHYSICS-CONNECT ACCESS
COLLEGE PHYSICS-CONNECT ACCESS
5th Edition
ISBN: 9781260486834
Author: GIAMBATTISTA
Publisher: MCG
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Chapter 3, Problem 51P
To determine

The velocity of the particle

Expert Solution & Answer
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Answer to Problem 51P

The velocity of the particle is 44.72m/s 26.56° south-east direction.

Explanation of Solution

Write the expression for the acceleration component due south.

    asouth=acosθ        (I)

Here, asouth is the acceleration in the south direction, a is the magnitude of acceleration and θ is the angle that velocity makes with the horizontal component.

Write the expression for the acceleration component due to east.

    aeast=asinθ        (II)

Here, aeast is the acceleration in the east direction, a is the magnitude of acceleration and θ is the angle that velocity makes with the vertical component.

Write the expression for the velocity at east direction after 8 seconds.

    veast=vi+aeastt        (III)

Here, veast is the velocity in the east direction, vi is the initial velocity, aeast is the acceleration in the east direction and t is the time.

Write the expression for the velocity at south direction after 8 seconds.

    vsouth=vi+asoutht        (IV)

Here, vsouth is the velocity in the south direction, vi is the initial velocity, asouth is the acceleration in the south direction and t is the time.

Write the equation for the speed after 8 seconds.

    v=(veast)2+(vsouth)2        (V)

Here, v is the speed after 8 seconds, veast is the velocity in the east direction and vsouth is the velocity in the south direction.

Write the expression to find the direction of the velocity,

  θ=tan1(vsouthvnorth)        (VI)

Here, θ is the direction, vsouth is the velocity in south, and vnorth is velocity in north

Conclusion;

Substitute 2.50m/s2 for a and 0° for θ in equation (I).

    asouth=(2.50m/s2)cos0°=2.50m/s2

Substitute 2.50m/s2 for a and 0° for θ in equation (II).

    aeast=(2.50m/s2)sin0°=0

Substitute 40m/s for vi, 0 for aeast in equation (III).

    veast=40m/s+(0)t=40m/s

Substitute 0m/s for vi, 2.50m/s2 for asouth and 8s for t in equation (IV).

    vsouth=0m/s+(2.50m/s2)(8s)=20m/s

Substitute 40m/s for veast and 20m/s for vsouth in equation (V).

    v=(40m/s)2+(20m/s)2=44.72m/s

Substitute 40m/s for veast and 20m/s for vsouth in equation (VI).

  θ=tan1(20m/s40m/s)=26.56°

Therefore, the velocity of the particle is 44.72m/s  26.56° south-east direction.

44.72m/s.

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Chapter 3 Solutions

COLLEGE PHYSICS-CONNECT ACCESS

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