The motion of a vibrating particle is defined by the position vector r = 10 ( 1 − e − 3 t ) i + ( 4 e − 2 t sin 15 t ) j , where r and t are expressed in millimeters and seconds, respectively. Determine the velocity and acceleration when (a) t = 0 , (b) t = 0.5 s.

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Answer 1

Textbook Question

Chapter 11.4, Problem 11.90P

The motion of a vibrating particle is defined by the position vector r = 10 ( 1 − e − 3 t ) i + ( 4 e − 2 t sin 15 t ) j , where r and t are expressed in millimeters and seconds, respectively. Determine the velocity and acceleration when (a) t = 0 , (b) t = 0.5 s.

Chapter 11.4, Problem 11.90P, The motion of a vibrating particle is defined by the position vector r=10(1e3t)i+(4e2tsin15t)j ,

Expert Solution

To determine

(a)

The velocity and acceleration when t=0.

Answer to Problem 11.90P

v=67.1 mm/s

a=256 mm/s2

Explanation of Solution

Given information:

The motion of a vibrating particle is defined by the position vector,

Vector Mechanics for Engineers: Dynamics, Chapter 11.4, Problem 11.90P , additional homework tip 1

r=10(1−e−3t)i+4(e−2tsin15t)j

Where, (r) in millimetres and (t) is in seconds.

We can obtain the velocity (v) at any time (t) by differentiating (r) with respect to (t),

Since, v=drdt

v=ddt(10(1−e−3t)i+(4e−2tsin15t)j)v= 30e−3ti+[60e−2tcos15t−8e−2tsin15t]j

The acceleration (a) can be obtained by differentiating again the above equation with respect to (t),

a=dvdta=−90e−3ti+[−120e−2tcos15t−900e−2tsin15t−120e−2tcos15t+16e−2tsin15t]ja=−90e−3ti+[−240e−2tcos15t−884e−2tsin15t]j

When t=0,

Velocity, v= 30e−3ti+[60e−2tcos15t−8e−2tsin15t]j

v= 30e−3(0)i+[60e−2(0)cos15(0)−8e−2(0)sin15(0)]jv= 30i+60j or,v=67.1 mm/s

And, acceleration, a=−90e−3ti+[−240e−2tcos15t−884e−2tsin15t]j

a=−90e−3(0)i+[−240e−2(0)cos15(0)−884e−2(0)sin15(0)]ja=−90i−240ja=256 mm/s2

Expert Solution

To determine

(b)

The velocity and acceleration when t=0.5.

Answer to Problem 11.90P

v=8.29 mm/s

a=336 mm/s2

P=173.2Ν

Explanation of Solution

Given information:

The motion of a vibrating particle is defined by the position vector,

Vector Mechanics for Engineers: Dynamics, Chapter 11.4, Problem 11.90P , additional homework tip 2

r=10(1−e−3t)i+4(e−2tsin15t)j

Where, (r) in millimetres and (t) is in seconds.

We can obtain the velocity (v) at any time (t) by differentiating (r) with respect to (t),

Since, v=drdt

v=ddt(10(1−e−3t)i+(4e−2tsin15t)j)v= 30e−3ti+[60e−2tcos15t−8e−2tsin15t]j

The acceleration (a) can be obtained by differentiating again the above equation with respect to (t),

a=dvdta=−90e−3ti+[−120e−2tcos15t−900e−2tsin15t−120e−2tcos15t+16e−2tsin15t]ja=−90e−3ti+[−240e−2tcos15t−884e−2tsin15t]j

When t=0.5,

Velocity, v= 30e−3ti+[60e−2tcos15t−8e−2tsin15t]j

v= 30e−3(0.5)i+[60e−2(0.5)cos15(0.5)−8e−2(0.5)sin15(0.5)]jv= 30e(−1.5)i+[60e−1cos(7.5)−8e−1sin(7.5)]jv= 6.694i+4.8906j or,v=8.29 mm/s

And, acceleration, a=−90e−3ti+[−240e−2tcos15t−884e−2tsin15t]j

a=−90e−3(0.5)i+[−240e−2(0.5)cos15(0.5)−884e−2(0.5)sin15(0.5)]ja=−90e(−1.5)i+[−240e−1cos(7.5)−884e−1sin(7.5)]ja=−20.08i−335.65ja=336 mm/s2

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Dynamics of Rigid Bodies problem:

3. The spatial motion of a particle is described by X = 3t 2 + 4t y = -4t 2 + 3t Z = -6t +9 where the coordinates are measured in feet and the time t is in seconds. (a) Determine the velocity and acceleration vectors of the particle as functions of time. (b) Verify that the particle is undergoing plane motion (the motion is not in a coordinate plane) by showing that the unit vector perpendicular to the plane formed by v and a is constant.

Answer 2

Textbook Question

Chapter 11.4, Problem 11.90P

The motion of a vibrating particle is defined by the position vector r = 10 ( 1 − e − 3 t ) i + ( 4 e − 2 t sin 15 t ) j , where r and t are expressed in millimeters and seconds, respectively. Determine the velocity and acceleration when (a) t = 0 , (b) t = 0.5 s.

Chapter 11.4, Problem 11.90P, The motion of a vibrating particle is defined by the position vector r=10(1e3t)i+(4e2tsin15t)j ,

Expert Solution

To determine

(a)

The velocity and acceleration when t=0.

Answer to Problem 11.90P

v=67.1 mm/s

a=256 mm/s2

Explanation of Solution

Given information:

The motion of a vibrating particle is defined by the position vector,

Vector Mechanics for Engineers: Dynamics, Chapter 11.4, Problem 11.90P , additional homework tip 1

r=10(1−e−3t)i+4(e−2tsin15t)j

Where, (r) in millimetres and (t) is in seconds.

We can obtain the velocity (v) at any time (t) by differentiating (r) with respect to (t),

Since, v=drdt

v=ddt(10(1−e−3t)i+(4e−2tsin15t)j)v= 30e−3ti+[60e−2tcos15t−8e−2tsin15t]j

The acceleration (a) can be obtained by differentiating again the above equation with respect to (t),

a=dvdta=−90e−3ti+[−120e−2tcos15t−900e−2tsin15t−120e−2tcos15t+16e−2tsin15t]ja=−90e−3ti+[−240e−2tcos15t−884e−2tsin15t]j

When t=0,

Velocity, v= 30e−3ti+[60e−2tcos15t−8e−2tsin15t]j

v= 30e−3(0)i+[60e−2(0)cos15(0)−8e−2(0)sin15(0)]jv= 30i+60j or,v=67.1 mm/s

And, acceleration, a=−90e−3ti+[−240e−2tcos15t−884e−2tsin15t]j

a=−90e−3(0)i+[−240e−2(0)cos15(0)−884e−2(0)sin15(0)]ja=−90i−240ja=256 mm/s2

Expert Solution

To determine

(b)

The velocity and acceleration when t=0.5.

Answer to Problem 11.90P

v=8.29 mm/s

a=336 mm/s2

P=173.2Ν

Explanation of Solution

Given information:

The motion of a vibrating particle is defined by the position vector,

Vector Mechanics for Engineers: Dynamics, Chapter 11.4, Problem 11.90P , additional homework tip 2

r=10(1−e−3t)i+4(e−2tsin15t)j

Where, (r) in millimetres and (t) is in seconds.

We can obtain the velocity (v) at any time (t) by differentiating (r) with respect to (t),

Since, v=drdt

v=ddt(10(1−e−3t)i+(4e−2tsin15t)j)v= 30e−3ti+[60e−2tcos15t−8e−2tsin15t]j

The acceleration (a) can be obtained by differentiating again the above equation with respect to (t),

a=dvdta=−90e−3ti+[−120e−2tcos15t−900e−2tsin15t−120e−2tcos15t+16e−2tsin15t]ja=−90e−3ti+[−240e−2tcos15t−884e−2tsin15t]j

When t=0.5,

Velocity, v= 30e−3ti+[60e−2tcos15t−8e−2tsin15t]j

v= 30e−3(0.5)i+[60e−2(0.5)cos15(0.5)−8e−2(0.5)sin15(0.5)]jv= 30e(−1.5)i+[60e−1cos(7.5)−8e−1sin(7.5)]jv= 6.694i+4.8906j or,v=8.29 mm/s

And, acceleration, a=−90e−3ti+[−240e−2tcos15t−884e−2tsin15t]j

a=−90e−3(0.5)i+[−240e−2(0.5)cos15(0.5)−884e−2(0.5)sin15(0.5)]ja=−90e(−1.5)i+[−240e−1cos(7.5)−884e−1sin(7.5)]ja=−20.08i−335.65ja=336 mm/s2

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Answer 3

Textbook Question

Chapter 11.4, Problem 11.90P

The motion of a vibrating particle is defined by the position vector r = 10 ( 1 − e − 3 t ) i + ( 4 e − 2 t sin 15 t ) j , where r and t are expressed in millimeters and seconds, respectively. Determine the velocity and acceleration when (a) t = 0 , (b) t = 0.5 s.

Chapter 11.4, Problem 11.90P, The motion of a vibrating particle is defined by the position vector r=10(1e3t)i+(4e2tsin15t)j ,

Expert Solution

To determine

(a)

The velocity and acceleration when t=0.

Answer to Problem 11.90P

v=67.1 mm/s

a=256 mm/s2

Explanation of Solution

Given information:

The motion of a vibrating particle is defined by the position vector,

Vector Mechanics for Engineers: Dynamics, Chapter 11.4, Problem 11.90P , additional homework tip 1

r=10(1−e−3t)i+4(e−2tsin15t)j

Where, (r) in millimetres and (t) is in seconds.

We can obtain the velocity (v) at any time (t) by differentiating (r) with respect to (t),

Since, v=drdt

v=ddt(10(1−e−3t)i+(4e−2tsin15t)j)v= 30e−3ti+[60e−2tcos15t−8e−2tsin15t]j

The acceleration (a) can be obtained by differentiating again the above equation with respect to (t),

a=dvdta=−90e−3ti+[−120e−2tcos15t−900e−2tsin15t−120e−2tcos15t+16e−2tsin15t]ja=−90e−3ti+[−240e−2tcos15t−884e−2tsin15t]j

When t=0,

Velocity, v= 30e−3ti+[60e−2tcos15t−8e−2tsin15t]j

v= 30e−3(0)i+[60e−2(0)cos15(0)−8e−2(0)sin15(0)]jv= 30i+60j or,v=67.1 mm/s

And, acceleration, a=−90e−3ti+[−240e−2tcos15t−884e−2tsin15t]j

a=−90e−3(0)i+[−240e−2(0)cos15(0)−884e−2(0)sin15(0)]ja=−90i−240ja=256 mm/s2

Expert Solution

To determine

(b)

The velocity and acceleration when t=0.5.

Answer to Problem 11.90P

v=8.29 mm/s

a=336 mm/s2

P=173.2Ν

Explanation of Solution

Given information:

The motion of a vibrating particle is defined by the position vector,

Vector Mechanics for Engineers: Dynamics, Chapter 11.4, Problem 11.90P , additional homework tip 2

r=10(1−e−3t)i+4(e−2tsin15t)j

Where, (r) in millimetres and (t) is in seconds.

We can obtain the velocity (v) at any time (t) by differentiating (r) with respect to (t),

Since, v=drdt

v=ddt(10(1−e−3t)i+(4e−2tsin15t)j)v= 30e−3ti+[60e−2tcos15t−8e−2tsin15t]j

The acceleration (a) can be obtained by differentiating again the above equation with respect to (t),

a=dvdta=−90e−3ti+[−120e−2tcos15t−900e−2tsin15t−120e−2tcos15t+16e−2tsin15t]ja=−90e−3ti+[−240e−2tcos15t−884e−2tsin15t]j

When t=0.5,

Velocity, v= 30e−3ti+[60e−2tcos15t−8e−2tsin15t]j

v= 30e−3(0.5)i+[60e−2(0.5)cos15(0.5)−8e−2(0.5)sin15(0.5)]jv= 30e(−1.5)i+[60e−1cos(7.5)−8e−1sin(7.5)]jv= 6.694i+4.8906j or,v=8.29 mm/s

And, acceleration, a=−90e−3ti+[−240e−2tcos15t−884e−2tsin15t]j

a=−90e−3(0.5)i+[−240e−2(0.5)cos15(0.5)−884e−2(0.5)sin15(0.5)]ja=−90e(−1.5)i+[−240e−1cos(7.5)−884e−1sin(7.5)]ja=−20.08i−335.65ja=336 mm/s2

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Answer 4

Textbook Question

Chapter 11.4, Problem 11.90P

The motion of a vibrating particle is defined by the position vector r = 10 ( 1 − e − 3 t ) i + ( 4 e − 2 t sin 15 t ) j , where r and t are expressed in millimeters and seconds, respectively. Determine the velocity and acceleration when (a) t = 0 , (b) t = 0.5 s.

Chapter 11.4, Problem 11.90P, The motion of a vibrating particle is defined by the position vector r=10(1e3t)i+(4e2tsin15t)j ,

Expert Solution

To determine

(a)

The velocity and acceleration when t=0.

Answer to Problem 11.90P

v=67.1 mm/s

a=256 mm/s2

Explanation of Solution

Given information:

The motion of a vibrating particle is defined by the position vector,

Vector Mechanics for Engineers: Dynamics, Chapter 11.4, Problem 11.90P , additional homework tip 1

r=10(1−e−3t)i+4(e−2tsin15t)j

Where, (r) in millimetres and (t) is in seconds.

We can obtain the velocity (v) at any time (t) by differentiating (r) with respect to (t),

Since, v=drdt

v=ddt(10(1−e−3t)i+(4e−2tsin15t)j)v= 30e−3ti+[60e−2tcos15t−8e−2tsin15t]j

The acceleration (a) can be obtained by differentiating again the above equation with respect to (t),

a=dvdta=−90e−3ti+[−120e−2tcos15t−900e−2tsin15t−120e−2tcos15t+16e−2tsin15t]ja=−90e−3ti+[−240e−2tcos15t−884e−2tsin15t]j

When t=0,

Velocity, v= 30e−3ti+[60e−2tcos15t−8e−2tsin15t]j

v= 30e−3(0)i+[60e−2(0)cos15(0)−8e−2(0)sin15(0)]jv= 30i+60j or,v=67.1 mm/s

And, acceleration, a=−90e−3ti+[−240e−2tcos15t−884e−2tsin15t]j

a=−90e−3(0)i+[−240e−2(0)cos15(0)−884e−2(0)sin15(0)]ja=−90i−240ja=256 mm/s2

Expert Solution

To determine

(b)

The velocity and acceleration when t=0.5.

Answer to Problem 11.90P

v=8.29 mm/s

a=336 mm/s2

P=173.2Ν

Explanation of Solution

Given information:

The motion of a vibrating particle is defined by the position vector,

Vector Mechanics for Engineers: Dynamics, Chapter 11.4, Problem 11.90P , additional homework tip 2

r=10(1−e−3t)i+4(e−2tsin15t)j

Where, (r) in millimetres and (t) is in seconds.

We can obtain the velocity (v) at any time (t) by differentiating (r) with respect to (t),

Since, v=drdt

v=ddt(10(1−e−3t)i+(4e−2tsin15t)j)v= 30e−3ti+[60e−2tcos15t−8e−2tsin15t]j

The acceleration (a) can be obtained by differentiating again the above equation with respect to (t),

a=dvdta=−90e−3ti+[−120e−2tcos15t−900e−2tsin15t−120e−2tcos15t+16e−2tsin15t]ja=−90e−3ti+[−240e−2tcos15t−884e−2tsin15t]j

When t=0.5,

Velocity, v= 30e−3ti+[60e−2tcos15t−8e−2tsin15t]j

v= 30e−3(0.5)i+[60e−2(0.5)cos15(0.5)−8e−2(0.5)sin15(0.5)]jv= 30e(−1.5)i+[60e−1cos(7.5)−8e−1sin(7.5)]jv= 6.694i+4.8906j or,v=8.29 mm/s

And, acceleration, a=−90e−3ti+[−240e−2tcos15t−884e−2tsin15t]j

a=−90e−3(0.5)i+[−240e−2(0.5)cos15(0.5)−884e−2(0.5)sin15(0.5)]ja=−90e(−1.5)i+[−240e−1cos(7.5)−884e−1sin(7.5)]ja=−20.08i−335.65ja=336 mm/s2

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Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

Expert Solution

To determine

(a)

The velocity and acceleration when t=0.

Answer to Problem 11.90P

v=67.1 mm/s

a=256 mm/s2

Explanation of Solution

Given information:

The motion of a vibrating particle is defined by the position vector,

Vector Mechanics for Engineers: Dynamics, Chapter 11.4, Problem 11.90P , additional homework tip 1

r=10(1−e−3t)i+4(e−2tsin15t)j

Where, (r) in millimetres and (t) is in seconds.

We can obtain the velocity (v) at any time (t) by differentiating (r) with respect to (t),

Since, v=drdt

v=ddt(10(1−e−3t)i+(4e−2tsin15t)j)v= 30e−3ti+[60e−2tcos15t−8e−2tsin15t]j

The acceleration (a) can be obtained by differentiating again the above equation with respect to (t),

a=dvdta=−90e−3ti+[−120e−2tcos15t−900e−2tsin15t−120e−2tcos15t+16e−2tsin15t]ja=−90e−3ti+[−240e−2tcos15t−884e−2tsin15t]j

When t=0,

Velocity, v= 30e−3ti+[60e−2tcos15t−8e−2tsin15t]j

v= 30e−3(0)i+[60e−2(0)cos15(0)−8e−2(0)sin15(0)]jv= 30i+60j or,v=67.1 mm/s

And, acceleration, a=−90e−3ti+[−240e−2tcos15t−884e−2tsin15t]j

a=−90e−3(0)i+[−240e−2(0)cos15(0)−884e−2(0)sin15(0)]ja=−90i−240ja=256 mm/s2

Expert Solution

To determine

(b)

The velocity and acceleration when t=0.5.

Answer to Problem 11.90P

v=8.29 mm/s

a=336 mm/s2

P=173.2Ν

Explanation of Solution

Given information:

The motion of a vibrating particle is defined by the position vector,

Vector Mechanics for Engineers: Dynamics, Chapter 11.4, Problem 11.90P , additional homework tip 2

r=10(1−e−3t)i+4(e−2tsin15t)j

Where, (r) in millimetres and (t) is in seconds.

We can obtain the velocity (v) at any time (t) by differentiating (r) with respect to (t),

Since, v=drdt

v=ddt(10(1−e−3t)i+(4e−2tsin15t)j)v= 30e−3ti+[60e−2tcos15t−8e−2tsin15t]j

The acceleration (a) can be obtained by differentiating again the above equation with respect to (t),

a=dvdta=−90e−3ti+[−120e−2tcos15t−900e−2tsin15t−120e−2tcos15t+16e−2tsin15t]ja=−90e−3ti+[−240e−2tcos15t−884e−2tsin15t]j

When t=0.5,

Velocity, v= 30e−3ti+[60e−2tcos15t−8e−2tsin15t]j

v= 30e−3(0.5)i+[60e−2(0.5)cos15(0.5)−8e−2(0.5)sin15(0.5)]jv= 30e(−1.5)i+[60e−1cos(7.5)−8e−1sin(7.5)]jv= 6.694i+4.8906j or,v=8.29 mm/s

And, acceleration, a=−90e−3ti+[−240e−2tcos15t−884e−2tsin15t]j

a=−90e−3(0.5)i+[−240e−2(0.5)cos15(0.5)−884e−2(0.5)sin15(0.5)]ja=−90e(−1.5)i+[−240e−1cos(7.5)−884e−1sin(7.5)]ja=−20.08i−335.65ja=336 mm/s2

Answer 8

Expert Solution

To determine

(a)

The velocity and acceleration when t=0.

Answer to Problem 11.90P

v=67.1 mm/s

a=256 mm/s2

Explanation of Solution

Given information:

The motion of a vibrating particle is defined by the position vector,

Vector Mechanics for Engineers: Dynamics, Chapter 11.4, Problem 11.90P , additional homework tip 1

r=10(1−e−3t)i+4(e−2tsin15t)j

Where, (r) in millimetres and (t) is in seconds.

We can obtain the velocity (v) at any time (t) by differentiating (r) with respect to (t),

Since, v=drdt

v=ddt(10(1−e−3t)i+(4e−2tsin15t)j)v= 30e−3ti+[60e−2tcos15t−8e−2tsin15t]j

The acceleration (a) can be obtained by differentiating again the above equation with respect to (t),

a=dvdta=−90e−3ti+[−120e−2tcos15t−900e−2tsin15t−120e−2tcos15t+16e−2tsin15t]ja=−90e−3ti+[−240e−2tcos15t−884e−2tsin15t]j

When t=0,

Velocity, v= 30e−3ti+[60e−2tcos15t−8e−2tsin15t]j

v= 30e−3(0)i+[60e−2(0)cos15(0)−8e−2(0)sin15(0)]jv= 30i+60j or,v=67.1 mm/s

And, acceleration, a=−90e−3ti+[−240e−2tcos15t−884e−2tsin15t]j

a=−90e−3(0)i+[−240e−2(0)cos15(0)−884e−2(0)sin15(0)]ja=−90i−240ja=256 mm/s2

Answer 9

Expert Solution

Answer 10

Expert Solution

Answer 11

Expert Solution

Answer 12

To determine

(a)

The velocity and acceleration when t=0.

Answer 13

Answer to Problem 11.90P

v=67.1 mm/s

a=256 mm/s2

Answer 14

Explanation of Solution

Given information:

The motion of a vibrating particle is defined by the position vector,

Vector Mechanics for Engineers: Dynamics, Chapter 11.4, Problem 11.90P , additional homework tip 1

r=10(1−e−3t)i+4(e−2tsin15t)j

Where, (r) in millimetres and (t) is in seconds.

We can obtain the velocity (v) at any time (t) by differentiating (r) with respect to (t),

Since, v=drdt

v=ddt(10(1−e−3t)i+(4e−2tsin15t)j)v= 30e−3ti+[60e−2tcos15t−8e−2tsin15t]j

The acceleration (a) can be obtained by differentiating again the above equation with respect to (t),

a=dvdta=−90e−3ti+[−120e−2tcos15t−900e−2tsin15t−120e−2tcos15t+16e−2tsin15t]ja=−90e−3ti+[−240e−2tcos15t−884e−2tsin15t]j

When t=0,

Velocity, v= 30e−3ti+[60e−2tcos15t−8e−2tsin15t]j

v= 30e−3(0)i+[60e−2(0)cos15(0)−8e−2(0)sin15(0)]jv= 30i+60j or,v=67.1 mm/s

And, acceleration, a=−90e−3ti+[−240e−2tcos15t−884e−2tsin15t]j

a=−90e−3(0)i+[−240e−2(0)cos15(0)−884e−2(0)sin15(0)]ja=−90i−240ja=256 mm/s2

Answer 15

Expert Solution

To determine

(b)

The velocity and acceleration when t=0.5.

Answer to Problem 11.90P

v=8.29 mm/s

a=336 mm/s2

P=173.2Ν

Explanation of Solution

Given information:

The motion of a vibrating particle is defined by the position vector,

Vector Mechanics for Engineers: Dynamics, Chapter 11.4, Problem 11.90P , additional homework tip 2

r=10(1−e−3t)i+4(e−2tsin15t)j

Where, (r) in millimetres and (t) is in seconds.

We can obtain the velocity (v) at any time (t) by differentiating (r) with respect to (t),

Since, v=drdt

v=ddt(10(1−e−3t)i+(4e−2tsin15t)j)v= 30e−3ti+[60e−2tcos15t−8e−2tsin15t]j

The acceleration (a) can be obtained by differentiating again the above equation with respect to (t),

a=dvdta=−90e−3ti+[−120e−2tcos15t−900e−2tsin15t−120e−2tcos15t+16e−2tsin15t]ja=−90e−3ti+[−240e−2tcos15t−884e−2tsin15t]j

When t=0.5,

Velocity, v= 30e−3ti+[60e−2tcos15t−8e−2tsin15t]j

v= 30e−3(0.5)i+[60e−2(0.5)cos15(0.5)−8e−2(0.5)sin15(0.5)]jv= 30e(−1.5)i+[60e−1cos(7.5)−8e−1sin(7.5)]jv= 6.694i+4.8906j or,v=8.29 mm/s

And, acceleration, a=−90e−3ti+[−240e−2tcos15t−884e−2tsin15t]j

a=−90e−3(0.5)i+[−240e−2(0.5)cos15(0.5)−884e−2(0.5)sin15(0.5)]ja=−90e(−1.5)i+[−240e−1cos(7.5)−884e−1sin(7.5)]ja=−20.08i−335.65ja=336 mm/s2

Answer 16

Expert Solution

Answer 17

Expert Solution

Answer 18

Expert Solution

Answer 19

To determine

(b)

The velocity and acceleration when t=0.5.

Answer 20

Answer to Problem 11.90P

v=8.29 mm/s

a=336 mm/s2

P=173.2Ν

Answer 21

Explanation of Solution

Given information:

The motion of a vibrating particle is defined by the position vector,

Vector Mechanics for Engineers: Dynamics, Chapter 11.4, Problem 11.90P , additional homework tip 2

r=10(1−e−3t)i+4(e−2tsin15t)j

Where, (r) in millimetres and (t) is in seconds.

We can obtain the velocity (v) at any time (t) by differentiating (r) with respect to (t),

Since, v=drdt

v=ddt(10(1−e−3t)i+(4e−2tsin15t)j)v= 30e−3ti+[60e−2tcos15t−8e−2tsin15t]j

The acceleration (a) can be obtained by differentiating again the above equation with respect to (t),

a=dvdta=−90e−3ti+[−120e−2tcos15t−900e−2tsin15t−120e−2tcos15t+16e−2tsin15t]ja=−90e−3ti+[−240e−2tcos15t−884e−2tsin15t]j

When t=0.5,

Velocity, v= 30e−3ti+[60e−2tcos15t−8e−2tsin15t]j

v= 30e−3(0.5)i+[60e−2(0.5)cos15(0.5)−8e−2(0.5)sin15(0.5)]jv= 30e(−1.5)i+[60e−1cos(7.5)−8e−1sin(7.5)]jv= 6.694i+4.8906j or,v=8.29 mm/s

And, acceleration, a=−90e−3ti+[−240e−2tcos15t−884e−2tsin15t]j

a=−90e−3(0.5)i+[−240e−2(0.5)cos15(0.5)−884e−2(0.5)sin15(0.5)]ja=−90e(−1.5)i+[−240e−1cos(7.5)−884e−1sin(7.5)]ja=−20.08i−335.65ja=336 mm/s2

Answer 22

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The motion of a vibrating particle is defined by the position vector r = 10 ( 1 − e − 3 t ) i + ( 4 e − 2 t sin 15 t ) j , where r and t are expressed in millimeters and seconds, respectively. Determine the velocity and acceleration when (a) t = 0 , (b) t = 0.5 s.

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Answer to Problem 11.90P

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Answer to Problem 11.90P

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