Principles of Physics: A Calculus-Based Text
Principles of Physics: A Calculus-Based Text
5th Edition
ISBN: 9781133104261
Author: Raymond A. Serway, John W. Jewett
Publisher: Cengage Learning
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Chapter 4, Problem 5P

(a)

To determine

The components of the particle’s velocity at t=10.0sec.

(a)

Expert Solution
Check Mark

Answer to Problem 5P

The x components of the particle’s velocity at t=10.0sec is 45m/s and the y component of particle’s velocity at t=10s is 15m/s.

Explanation of Solution

Formula to calculate net force acts on a particle is,

    Fnet=F1+F2

Here, Fnet is the net force acting on a particle, F1 and F2 are two forces on the particle.

Formula to calculate acceleration of a particle is,

    a=Fnetm

Here, a is the acceleration of the particle and m is the mass of the particle.

Substitute F1+F2 for Fnet in above equation.

    a=F1+F2m

Substitute (6.00i^4.00j^)N for F1, (3.00i^+7.00j^)N for F2 and 2.00kg for m to find a.

    a=(6.00i^4.00j^)N+(3.00i^+7.00j^)N2.00kg=(4.50m/s2)i^+(1.50m/s2)j^

Formula to calculate the velocity of the particle is,

    vf=vi+at

Here, vf is the final velocity of the particle, vi is the initial velocity, and t is time.

Substitute 0m/s for vi, (4.5i^+1.5j^)m/s2 for a and 10.0s for t to find vf.

    vf=0+[(4.5i^+1.5j^)m/s2×(10.0s)]=(45i^+15j^)m/s

Conclusion:

Therefore, the x components of the particle’s velocity at t=10.0sec is 45m/s and the y component of particle’s velocity at t=10s is 15m/s.

(b)

To determine

The direction of the motion of the particle at t=10.0sec.

(b)

Expert Solution
Check Mark

Answer to Problem 5P

The direction of the particle’s motion at t=10.0sec is 162° counterclockwise from the +x axis.

Explanation of Solution

Formula to calculate the direction of the moving particle is,

    θ=tan1(vyvx)

Here, θ is the direction of the velocity of the particle, vx and vy are horizontal and vertical components of the particle’s velocity.

Substitute 15m/s for vy and 45m/s for vx to calculate θ.

    θ=tan1(15m/s45m/s)θ162°

Conclusion:

Therefore, the direction of particle’s motion at t=10.0sec is 162° counterclockwise from the +x axis.

(c)

To determine

The displacement of the particle during 10sec.

(c)

Expert Solution
Check Mark

Answer to Problem 5P

The displacement of the particle during 10sec is (225i^+75j^)m.

Explanation of Solution

Formula to calculate the displacement of the particle is,

    s=vit+12at2

Here, s is the displacement of the particle.

Substitute 0m/s for vi, (4.5i^+1.5j^)m/s2 for a and 10.0s for t to find s.

    s=[0×(10.0s)]+[12(4.5i^+1.5j^)m/s2(10.0s)2]=(255i^+75j^)m

Conclusion:

Therefore, the displacement of the particle during 10sec is (255i^+75j^)m.

(d)

To determine

The coordinates of the particle at t=10.0sec.

(d)

Expert Solution
Check Mark

Answer to Problem 5P

The coordinates of the particle at t=10.0sec is (227i^+79j^)m.

Explanation of Solution

The initial position of the particle is (2i^+4j^)m.

The final coordinates of the particle at t=10.0s is,

    Final coordinates=(255i^+75j^)m+(2i^+4j^)m=(227i^+79j^)m

Conclusion:

Therefore, the coordinates of the particle at t=10.0sec is (227i^+79j^)m

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Chapter 4 Solutions

Principles of Physics: A Calculus-Based Text

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