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 3, Problem 3P

A particle initially located at the origin has an acceleration of a = 3.00 j ^ m/s 2 and an initial velocity of v i = 5.00 i ^ m/s . Find (a) the vector position of the particle at any time t, (b) the velocity of the particle at any time t, (c) the coordinates of the particle at t = 2.00 s, and (d) the speed of the particle at t = 2.00 s.

(a)

Expert Solution
Check Mark
To determine

The vector position of the particle at time t .

Answer to Problem 3P

The vector position of the particle at time t is 5.00ti^+1.50t2j^_.

Explanation of Solution

Write the expression for the vector position of the particle.

  rf=ri+vit+12at2        (I)

Here, rf is the final vector position of the particle, ri is the initial vector position of the particle, vi is the initial velocity of the particle, a is the acceleration of the particle and t is the time taken.

Conclusion:

Substitute 3.00j^m/s2 for a, 5.00i^ m/s for vi and 0i^+0j^ for ri in above equation to find rf,

  rf=(0i^+0j^)+(5.00i^ m/s)t+12(3.00j^m/s2)t2=5.00ti^+1.50t2j^

Therefore, the vector position of the particle at time t is 5.00ti^+1.50t2j^_.

(b)

Expert Solution
Check Mark
To determine

The velocity of the particle at any time t.

Answer to Problem 3P

The velocity of the particle at any time t is 5.00ti^+3.00tj^.

Explanation of Solution

Write the expression for the velocity of the particle at any time t.

  vf=vi+at        (II)

Here, vf is the velocity of the particle at any time t, vi is the initial velocity of the particle.

Conclusion:

Substitute 3.00j^m/s2 for a and 5.00i^ m/s for vi in above equation to find vf,

  vf=5.00i^+3.00tj^

Therefore, the velocity of the particle at any time t is 5.00ti^+3.00tj^.

(c)

Expert Solution
Check Mark
To determine

The coordinates of the particle at t=2.00s.

Answer to Problem 3P

The coordinates of the particle at t=2.00s is xf=10.0m, yf=6.00m.

Explanation of Solution

Write the expression for the vector position of the particle at time t.

  rf=xfi^+yfj^        (III)

Here, xf is the x component and yf is the y component.

Conclusion:

Substitute 2.0s for t a in rf=5.00ti^+1.50t2j^ equation to find xf and yf.

  rf=(5.00)(2.00s)i^+(1.50)(2.00s)2j^=10.00i^+6.00j^

So compare the above value with (III) to get xf and yf,

  xf=10.0m

  yf=6.0m

Therefore, the coordinates of the particle at t=2.00s is xf=10.0m, yf=6.00m.

(d)

Expert Solution
Check Mark
To determine

The speed of the particle at t=2.00s.

Answer to Problem 3P

The speed of the particle at t=2.00s is 7.81m/s.

Explanation of Solution

Write the expression for the speed of the particle at time t.

  vf=vi+at        (IV)

Write the final speed of the particle.

    vf=vxf2+vyf2        (V)

Here, vxf is the speed at x component and vyf is the speed at y component.

Conclusion:

Substitute 3.00j^m/s2 for a and 5.00i^ m/s for vi in (IV) to find vf,

  vf=(5.00m/s)i^+(3.00m/s)(2.00s)j^=(5.00i^+6.00j^)m/s

Substitute 5.00i^m/s for vxf and 6.00j^ m/s for vyf in (V) to find vf,

    vf=(5.00i^m/s)2+(6.00j^ m/s)2=7.81m/s

Therefore, the speed of the particle at t=2.00s is 7.81m/s.

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Chapter 3 Solutions

Principles of Physics: A Calculus-Based Text

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