Physics for Scientists and Engineers: Foundations and Connections
Physics for Scientists and Engineers: Foundations and Connections
1st Edition
ISBN: 9781133939146
Author: Katz, Debora M.
Publisher: Cengage Learning
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Chapter 4, Problem 11PQ

(a)

To determine

Write the position vector of the object as a function of time.

(a)

Expert Solution
Check Mark

Answer to Problem 11PQ

The position vector of the object as a function of time is (1.25t2i^+3.00tj^) m.

Explanation of Solution

An object is moving with a initial velocity vi=3.00j^ m/s and with acceleration a=2.50i^ m/s2 and the object was initially at the origin (0,0).

Write the formula for the position vector [two-dimensional kinematic equation]

    r=ri+vit+12at2 (I)

Here, r is the final position vector, ri is the initial position of the object, vi is the initial velocity, t is time and a is the acceleration.

Conclusion:

The initial position vector is ri=0i^+0j^ as the object was initially at the origin.

Substitute 3.00j^ m/s for vi, 0i^+0j^ for ri and 2.50i^ m/s2 for a in equation (I) to find the value of r

r=0i^+0j^+3.00j^ m/s×t+12(2.50i^ m/s2)t2r=(1.25t2i^+3.00tj^) m

Thus, the position vector of the object as a function of time is (1.25t2i^+3.00tj^) m.

(b)

To determine

Write the velocity vector of the object as a function of time.

(b)

Expert Solution
Check Mark

Answer to Problem 11PQ

The velocity vector of the object as a function of time is (2.50ti^+3.00j^) m/s.

Explanation of Solution

Write the formula for the velocity vector [two-dimensional kinematic equation]

    vf=vi+at (II)

Here, vf is the final velocity vector.

Conclusion:

Substitute 3.00j^ m/s for vi and 2.50i^ m/s2 for a in equation (II) to find the value of vf

vf=3.00j^ m/s+2.50i^ m/s2tvf=(2.50ti^+3.00j^) m/s

Thus, the velocity vector of the object as a function of time is (2.50ti^+3.00j^) m/s.

(c)

To determine

Write the position vector of the object at time t=3.00 s.

(c)

Expert Solution
Check Mark

Answer to Problem 11PQ

The position vector of the object at time t=3.00 s is (11.3i^+9.00j^) m.

Explanation of Solution

The position of the object after 3.00 s is

Substitute 3.00j^ m/s for vi, 0i^+0j^ for ri, 3.00 s for t and 2.50i^ m/s2 for a in equation (I) to find the value of r

r=0i^+0j^+3.00j^ m/s×3.00 s+12(2.50i^ m/s2)(3.00 s)2r=(1.25×(3.00)2i^+3.00×3.00j^) mr=(11.3i^+9.00j^) m

Thus, the position vector of the object at time t=3.00 s is (11.3i^+9.00j^) m.

(d)

To determine

Write the speed of the object at time t=3.00 s.

(d)

Expert Solution
Check Mark

Answer to Problem 11PQ

The speed of the object at time t=3.00 s is 8.08 m/s.

Explanation of Solution

The velocity of the object after 3.00 s is

Substitute 3.00j^ m/s for vi, 3.00 s for t and 2.50i^ m/s2 for a in equation (II) to find the value of vf

vf=3.00j^ m/s+2.50i^×3.00  m/svf=(7.50i^+3.00j^) m/s

Thus, the velocity vector of the object at time t=3.00 s is (7.50i^+3.00j^) m/s.

The speed of the object is vf=v2xf+v2yf=(7.50)2+(3.00)2=8.08 m/s.

Thus, the speed of the object at time t=3.00 s is 8.08 m/s.

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Chapter 4 Solutions

Physics for Scientists and Engineers: Foundations and Connections

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Position/Velocity/Acceleration Part 1: Definitions; Author: Professor Dave explains;https://www.youtube.com/watch?v=4dCrkp8qgLU;License: Standard YouTube License, CC-BY