A Solution Manual for Vector Mechanics for

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Solution Manual for Vector Mechanics for

동역학 / Dynamics (숭실대학교)

A StuDocu não é patrocinada ou endossada por nenhuma faculdade ou universidade

Baixado por Matheus Silva ([e-mail])

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CHAPTER 11

Full file at https://testbanku.eu/Solution-Manual-for-Vector-Mechanics-for-Engineers-Dynamics-11th-Edition-by-BeerBaixadoporMatheusSilva([e-mail])

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Full file at https://testbanku.eu/Solution-Manual-for-Vector-Mechanics-for-Engineers-Dynamics-11th-Edition-by-BeerBaixadoporMatheusSilva([e-mail])

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PROBLEM 11.1

A snowboarder starts from rest at the top of a double black diamond hill. As he rides down the slope, GPS coordinates are used to determine his displacement as a function of time: x= 0.5t + t + 2t where x and t are expressed in ft and seconds, respectively. Determine the position, velocity, and acceleration of the boarder when t = 5 seconds.

3 2

2

2

2

SOLUTION

Position:

---

x  0.5t  t  2t

Velocity:

---

v 

---

dx

dt

---

 1.5t  2t  2

Acceleration:

---

a 

---

dv

dt

---

 3t  2

At t  5 s,

---

x  0.5

---



---

5

---



---

3

---

 5  2

---



---

5

---



---

x  97.5 ft ◄

v  1.5

---



---

5

---



---

2

---

 2

---



---

5

---

 2 v  49.5 ft/s ◄

a  3

---



---

5

---



---

 2  

---

a  17 ft/s ◄

Copyright © McGraw-Hill Education. Permission required for reproduction or display.

Full file at https://testbanku.eu/Solution-Manual-for-Vector-Mechanics-for-Engineers-Dynamics-11th-Edition-by-BeerBaixadoporMatheusSilva([e-mail])

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PROBLEM 11.2

The motion of a particle is defined by the relation x  2t  9t  12t  10, where x and t are expressed in feet and seconds, respectively. Determine the time, the position, and the acceleration of the particle when v  0.

3 2

dx

2 2

1

1

1

2

3 2

2

2

2

2

SOLUTION

Differentiating,

So v  0 at t  1 s and t  2 s. At t  1 s,

---

x  2t  9t  12t  10

v   6t  18t  12  6(t  3t  2) dt

 6(t  2)(t  1)

dv

a 

 12t  18

dt

x1  2  9  12  10  15

a

 12  18  6

---

a

---

x

t  1.000 s ◄  15.00 ft ◄

 6.00 ft/s ◄

At t  2 s,

x2  2(2)  9(2)  12(2)  10  14

---

t  2.00 s ◄

a

---

 (12)(2)  18  6

---

a

---

x

 14.00 ft ◄

 6.00 ft/s ◄

Full file at https://testbanku.eu/Solution-Manual-for-Vector-Mechanics-for-Engineers-Dynamics-11th-Edition-by-BeerBaixadoporMatheusSilva([e-mail])