(a) The magnitude of the velocity of the parasailer as a function of time.

Question

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Answer 1

Question

Chapter 11.5, Problem 11.164P

To determine

(a)

The magnitude of the velocity of the parasailer as a function of time.

Expert Solution

Answer to Problem 11.164P

Vector Mechanics For Engineers, Chapter 11.5, Problem 11.164P , additional homework tip 1

Explanation of Solution

Given information:

Vector Mechanics For Engineers, Chapter 11.5, Problem 11.164P , additional homework tip 2

Length of rope is defined as,

$r=600−18t5/2$

Constant velocity of the boat is $15 kn$

The angle is increasing at $2°s$

The velocity in radial and transverse components,

$v=vrer+vθeθ=r˙er+rθ˙eθ$

The acceleration in radial and transverse components,

$a=arer+aθeθ=(r¨−rθ˙2)er+(rθ¨+2r˙θ˙)eθ$

Calculation:

Vector Mechanics For Engineers, Chapter 11.5, Problem 11.164P , additional homework tip 3

According to given information,

Convert,

$vB=15 kn=25.3 ft/saB=0$

We know that,

$r=600−18t5/2$

$θ˙=2°s =0.0349 rad/s$

Therefore,

$r˙=−516t3/2r¨=−1532t1/2$

$θ¨=0$

According to the explanation, the relative velocity of parasailer with respect to boat is,

$vP/B=r˙er+rθ˙eθ$

The relative co-ordinates of the relative velocity is equal to,

$vP/B=r˙(−cosθi+sinθj)+rθ˙(sinθi+cosθj)$

Therefore the velocity of parasailer is equal to,

$vP=vB+vP/BvP=vB+r˙(−cosθi+sinθj)+rθ˙(sinθi+cosθj)vP=(vB−r˙cosθ+rθ˙sinθ)i+(r˙sinθ+rθ˙cosθ)j$

We can rewrite this as,

$vP=vPxi+vPyj$

Where,

$vPx=vB−r˙cosθ+rθ˙sinθvPy=r˙sinθ+rθ˙cosθ$

To find the magnitude,

$vP=vPx2+vPy2$

Now, plot the graph,

Vector Mechanics For Engineers, Chapter 11.5, Problem 11.164P , additional homework tip 4

Conclusion:

The magnitude of velocity of the parasailer is equal to,

$vP=vPx2+vPy2$

The relevant graph is shown above.

To determine

(b)

The magnitude of acceleration of parasailer at $t=5 s$

Expert Solution

Answer to Problem 11.164P

$aP=1.7871 m/s2$

Explanation of Solution

Given information:

Vector Mechanics For Engineers, Chapter 11.5, Problem 11.164P , additional homework tip 5

Length of rope is defined as,

$r=600−18t5/2$

Constant velocity of the boat is $15 kn$

The angle is increasing at $2°s$

The velocity in radial and transverse components,

$v=vrer+vθeθ=r˙er+rθ˙eθ$

The acceleration in radial and transverse components,

$a=arer+aθeθ=(r¨−rθ˙2)er+(rθ¨+2r˙θ˙)eθ$

Calculation:

Vector Mechanics For Engineers, Chapter 11.5, Problem 11.164P , additional homework tip 6

According to given information,

Convert,

$vB=15 kn=25.3 ft/saB=0$

We know that,

$r=600−18t5/2$

$θ˙=2°s =0.0349 rad/s$

Therefore,

$r˙=−516t3/2r¨=−1532t1/2$

$θ¨=0$

According to the explanation, the relative acceleration of parasailer with respect to boat is,

$aP/B=(r¨−rθ˙2)er+(rθ¨+2r˙θ˙)eθaP/B=[−1532t1/2−(600−18t5/2)(0.0349 rad/s)2]er+[2(−516t3/2)(0.0349 rad/s)]eθ$

The acceleration of parasailer is equal to,

$aP=aB+aP/BaP=0+[(−1532t1/2−(600−18t5/2)(0.0349 rad/s)2)er+(2(−516t3/2)(0.0349 rad/s))eθ]$

At, $t=5 s$

$aP=[(−1532(5)1/2−(600−18(5)5/2)(0.0349 rad/s)2)er+(2(−516(5)3/2)(0.0349 rad/s))eθ]aP=−1.7704er−0.2439eθ$

Therefore the magnitude is equal to,

$aP=(−1.7704)2+(−0.2439)2aP=1.7871 m/s2$

Conclusion:

The magnitude of acceleration of the parasailer is equal to,

$aP=1.7871 m/s2$

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Answer 2

Question

Chapter 11.5, Problem 11.164P

To determine

(a)

The magnitude of the velocity of the parasailer as a function of time.

Expert Solution

Answer to Problem 11.164P

Vector Mechanics For Engineers, Chapter 11.5, Problem 11.164P , additional homework tip 1

Explanation of Solution

Given information:

Vector Mechanics For Engineers, Chapter 11.5, Problem 11.164P , additional homework tip 2

Length of rope is defined as,

$r=600−18t5/2$

Constant velocity of the boat is $15 kn$

The angle is increasing at $2°s$

The velocity in radial and transverse components,

$v=vrer+vθeθ=r˙er+rθ˙eθ$

The acceleration in radial and transverse components,

$a=arer+aθeθ=(r¨−rθ˙2)er+(rθ¨+2r˙θ˙)eθ$

Calculation:

Vector Mechanics For Engineers, Chapter 11.5, Problem 11.164P , additional homework tip 3

According to given information,

Convert,

$vB=15 kn=25.3 ft/saB=0$

We know that,

$r=600−18t5/2$

$θ˙=2°s =0.0349 rad/s$

Therefore,

$r˙=−516t3/2r¨=−1532t1/2$

$θ¨=0$

According to the explanation, the relative velocity of parasailer with respect to boat is,

$vP/B=r˙er+rθ˙eθ$

The relative co-ordinates of the relative velocity is equal to,

$vP/B=r˙(−cosθi+sinθj)+rθ˙(sinθi+cosθj)$

Therefore the velocity of parasailer is equal to,

$vP=vB+vP/BvP=vB+r˙(−cosθi+sinθj)+rθ˙(sinθi+cosθj)vP=(vB−r˙cosθ+rθ˙sinθ)i+(r˙sinθ+rθ˙cosθ)j$

We can rewrite this as,

$vP=vPxi+vPyj$

Where,

$vPx=vB−r˙cosθ+rθ˙sinθvPy=r˙sinθ+rθ˙cosθ$

To find the magnitude,

$vP=vPx2+vPy2$

Now, plot the graph,

Vector Mechanics For Engineers, Chapter 11.5, Problem 11.164P , additional homework tip 4

Conclusion:

The magnitude of velocity of the parasailer is equal to,

$vP=vPx2+vPy2$

The relevant graph is shown above.

To determine

(b)

The magnitude of acceleration of parasailer at $t=5 s$

Expert Solution

Answer to Problem 11.164P

$aP=1.7871 m/s2$

Explanation of Solution

Given information:

Vector Mechanics For Engineers, Chapter 11.5, Problem 11.164P , additional homework tip 5

Length of rope is defined as,

$r=600−18t5/2$

Constant velocity of the boat is $15 kn$

The angle is increasing at $2°s$

The velocity in radial and transverse components,

$v=vrer+vθeθ=r˙er+rθ˙eθ$

The acceleration in radial and transverse components,

$a=arer+aθeθ=(r¨−rθ˙2)er+(rθ¨+2r˙θ˙)eθ$

Calculation:

Vector Mechanics For Engineers, Chapter 11.5, Problem 11.164P , additional homework tip 6

According to given information,

Convert,

$vB=15 kn=25.3 ft/saB=0$

We know that,

$r=600−18t5/2$

$θ˙=2°s =0.0349 rad/s$

Therefore,

$r˙=−516t3/2r¨=−1532t1/2$

$θ¨=0$

According to the explanation, the relative acceleration of parasailer with respect to boat is,

$aP/B=(r¨−rθ˙2)er+(rθ¨+2r˙θ˙)eθaP/B=[−1532t1/2−(600−18t5/2)(0.0349 rad/s)2]er+[2(−516t3/2)(0.0349 rad/s)]eθ$

The acceleration of parasailer is equal to,

$aP=aB+aP/BaP=0+[(−1532t1/2−(600−18t5/2)(0.0349 rad/s)2)er+(2(−516t3/2)(0.0349 rad/s))eθ]$

At, $t=5 s$

$aP=[(−1532(5)1/2−(600−18(5)5/2)(0.0349 rad/s)2)er+(2(−516(5)3/2)(0.0349 rad/s))eθ]aP=−1.7704er−0.2439eθ$

Therefore the magnitude is equal to,

$aP=(−1.7704)2+(−0.2439)2aP=1.7871 m/s2$

Conclusion:

The magnitude of acceleration of the parasailer is equal to,

$aP=1.7871 m/s2$

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Answer 3

Question

Chapter 11.5, Problem 11.164P

To determine

(a)

The magnitude of the velocity of the parasailer as a function of time.

Expert Solution

Answer to Problem 11.164P

Vector Mechanics For Engineers, Chapter 11.5, Problem 11.164P , additional homework tip 1

Explanation of Solution

Given information:

Vector Mechanics For Engineers, Chapter 11.5, Problem 11.164P , additional homework tip 2

Length of rope is defined as,

$r=600−18t5/2$

Constant velocity of the boat is $15 kn$

The angle is increasing at $2°s$

The velocity in radial and transverse components,

$v=vrer+vθeθ=r˙er+rθ˙eθ$

The acceleration in radial and transverse components,

$a=arer+aθeθ=(r¨−rθ˙2)er+(rθ¨+2r˙θ˙)eθ$

Calculation:

Vector Mechanics For Engineers, Chapter 11.5, Problem 11.164P , additional homework tip 3

According to given information,

Convert,

$vB=15 kn=25.3 ft/saB=0$

We know that,

$r=600−18t5/2$

$θ˙=2°s =0.0349 rad/s$

Therefore,

$r˙=−516t3/2r¨=−1532t1/2$

$θ¨=0$

According to the explanation, the relative velocity of parasailer with respect to boat is,

$vP/B=r˙er+rθ˙eθ$

The relative co-ordinates of the relative velocity is equal to,

$vP/B=r˙(−cosθi+sinθj)+rθ˙(sinθi+cosθj)$

Therefore the velocity of parasailer is equal to,

$vP=vB+vP/BvP=vB+r˙(−cosθi+sinθj)+rθ˙(sinθi+cosθj)vP=(vB−r˙cosθ+rθ˙sinθ)i+(r˙sinθ+rθ˙cosθ)j$

We can rewrite this as,

$vP=vPxi+vPyj$

Where,

$vPx=vB−r˙cosθ+rθ˙sinθvPy=r˙sinθ+rθ˙cosθ$

To find the magnitude,

$vP=vPx2+vPy2$

Now, plot the graph,

Vector Mechanics For Engineers, Chapter 11.5, Problem 11.164P , additional homework tip 4

Conclusion:

The magnitude of velocity of the parasailer is equal to,

$vP=vPx2+vPy2$

The relevant graph is shown above.

To determine

(b)

The magnitude of acceleration of parasailer at $t=5 s$

Expert Solution

Answer to Problem 11.164P

$aP=1.7871 m/s2$

Explanation of Solution

Given information:

Vector Mechanics For Engineers, Chapter 11.5, Problem 11.164P , additional homework tip 5

Length of rope is defined as,

$r=600−18t5/2$

Constant velocity of the boat is $15 kn$

The angle is increasing at $2°s$

The velocity in radial and transverse components,

$v=vrer+vθeθ=r˙er+rθ˙eθ$

The acceleration in radial and transverse components,

$a=arer+aθeθ=(r¨−rθ˙2)er+(rθ¨+2r˙θ˙)eθ$

Calculation:

Vector Mechanics For Engineers, Chapter 11.5, Problem 11.164P , additional homework tip 6

According to given information,

Convert,

$vB=15 kn=25.3 ft/saB=0$

We know that,

$r=600−18t5/2$

$θ˙=2°s =0.0349 rad/s$

Therefore,

$r˙=−516t3/2r¨=−1532t1/2$

$θ¨=0$

According to the explanation, the relative acceleration of parasailer with respect to boat is,

$aP/B=(r¨−rθ˙2)er+(rθ¨+2r˙θ˙)eθaP/B=[−1532t1/2−(600−18t5/2)(0.0349 rad/s)2]er+[2(−516t3/2)(0.0349 rad/s)]eθ$

The acceleration of parasailer is equal to,

$aP=aB+aP/BaP=0+[(−1532t1/2−(600−18t5/2)(0.0349 rad/s)2)er+(2(−516t3/2)(0.0349 rad/s))eθ]$

At, $t=5 s$

$aP=[(−1532(5)1/2−(600−18(5)5/2)(0.0349 rad/s)2)er+(2(−516(5)3/2)(0.0349 rad/s))eθ]aP=−1.7704er−0.2439eθ$

Therefore the magnitude is equal to,

$aP=(−1.7704)2+(−0.2439)2aP=1.7871 m/s2$

Conclusion:

The magnitude of acceleration of the parasailer is equal to,

$aP=1.7871 m/s2$

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Answer 4

Question

Chapter 11.5, Problem 11.164P

To determine

(a)

The magnitude of the velocity of the parasailer as a function of time.

Expert Solution

Answer to Problem 11.164P

Vector Mechanics For Engineers, Chapter 11.5, Problem 11.164P , additional homework tip 1

Explanation of Solution

Given information:

Vector Mechanics For Engineers, Chapter 11.5, Problem 11.164P , additional homework tip 2

Length of rope is defined as,

$r=600−18t5/2$

Constant velocity of the boat is $15 kn$

The angle is increasing at $2°s$

The velocity in radial and transverse components,

$v=vrer+vθeθ=r˙er+rθ˙eθ$

The acceleration in radial and transverse components,

$a=arer+aθeθ=(r¨−rθ˙2)er+(rθ¨+2r˙θ˙)eθ$

Calculation:

Vector Mechanics For Engineers, Chapter 11.5, Problem 11.164P , additional homework tip 3

According to given information,

Convert,

$vB=15 kn=25.3 ft/saB=0$

We know that,

$r=600−18t5/2$

$θ˙=2°s =0.0349 rad/s$

Therefore,

$r˙=−516t3/2r¨=−1532t1/2$

$θ¨=0$

According to the explanation, the relative velocity of parasailer with respect to boat is,

$vP/B=r˙er+rθ˙eθ$

The relative co-ordinates of the relative velocity is equal to,

$vP/B=r˙(−cosθi+sinθj)+rθ˙(sinθi+cosθj)$

Therefore the velocity of parasailer is equal to,

$vP=vB+vP/BvP=vB+r˙(−cosθi+sinθj)+rθ˙(sinθi+cosθj)vP=(vB−r˙cosθ+rθ˙sinθ)i+(r˙sinθ+rθ˙cosθ)j$

We can rewrite this as,

$vP=vPxi+vPyj$

Where,

$vPx=vB−r˙cosθ+rθ˙sinθvPy=r˙sinθ+rθ˙cosθ$

To find the magnitude,

$vP=vPx2+vPy2$

Now, plot the graph,

Vector Mechanics For Engineers, Chapter 11.5, Problem 11.164P , additional homework tip 4

Conclusion:

The magnitude of velocity of the parasailer is equal to,

$vP=vPx2+vPy2$

The relevant graph is shown above.

To determine

(b)

The magnitude of acceleration of parasailer at $t=5 s$

Expert Solution

Answer to Problem 11.164P

$aP=1.7871 m/s2$

Explanation of Solution

Given information:

Vector Mechanics For Engineers, Chapter 11.5, Problem 11.164P , additional homework tip 5

Length of rope is defined as,

$r=600−18t5/2$

Constant velocity of the boat is $15 kn$

The angle is increasing at $2°s$

The velocity in radial and transverse components,

$v=vrer+vθeθ=r˙er+rθ˙eθ$

The acceleration in radial and transverse components,

$a=arer+aθeθ=(r¨−rθ˙2)er+(rθ¨+2r˙θ˙)eθ$

Calculation:

Vector Mechanics For Engineers, Chapter 11.5, Problem 11.164P , additional homework tip 6

According to given information,

Convert,

$vB=15 kn=25.3 ft/saB=0$

We know that,

$r=600−18t5/2$

$θ˙=2°s =0.0349 rad/s$

Therefore,

$r˙=−516t3/2r¨=−1532t1/2$

$θ¨=0$

According to the explanation, the relative acceleration of parasailer with respect to boat is,

$aP/B=(r¨−rθ˙2)er+(rθ¨+2r˙θ˙)eθaP/B=[−1532t1/2−(600−18t5/2)(0.0349 rad/s)2]er+[2(−516t3/2)(0.0349 rad/s)]eθ$

The acceleration of parasailer is equal to,

$aP=aB+aP/BaP=0+[(−1532t1/2−(600−18t5/2)(0.0349 rad/s)2)er+(2(−516t3/2)(0.0349 rad/s))eθ]$

At, $t=5 s$

$aP=[(−1532(5)1/2−(600−18(5)5/2)(0.0349 rad/s)2)er+(2(−516(5)3/2)(0.0349 rad/s))eθ]aP=−1.7704er−0.2439eθ$

Therefore the magnitude is equal to,

$aP=(−1.7704)2+(−0.2439)2aP=1.7871 m/s2$

Conclusion:

The magnitude of acceleration of the parasailer is equal to,

$aP=1.7871 m/s2$

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Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 5

Chapter 11.5, Problem 11.164P

To determine

(a)

The magnitude of the velocity of the parasailer as a function of time.

Expert Solution

Answer to Problem 11.164P

Vector Mechanics For Engineers, Chapter 11.5, Problem 11.164P , additional homework tip 1

Explanation of Solution

Given information:

Vector Mechanics For Engineers, Chapter 11.5, Problem 11.164P , additional homework tip 2

Length of rope is defined as,

$r=600−18t5/2$

Constant velocity of the boat is $15 kn$

The angle is increasing at $2°s$

The velocity in radial and transverse components,

$v=vrer+vθeθ=r˙er+rθ˙eθ$

The acceleration in radial and transverse components,

$a=arer+aθeθ=(r¨−rθ˙2)er+(rθ¨+2r˙θ˙)eθ$

Calculation:

Vector Mechanics For Engineers, Chapter 11.5, Problem 11.164P , additional homework tip 3

According to given information,

Convert,

$vB=15 kn=25.3 ft/saB=0$

We know that,

$r=600−18t5/2$

$θ˙=2°s =0.0349 rad/s$

Therefore,

$r˙=−516t3/2r¨=−1532t1/2$

$θ¨=0$

According to the explanation, the relative velocity of parasailer with respect to boat is,

$vP/B=r˙er+rθ˙eθ$

The relative co-ordinates of the relative velocity is equal to,

$vP/B=r˙(−cosθi+sinθj)+rθ˙(sinθi+cosθj)$

Therefore the velocity of parasailer is equal to,

$vP=vB+vP/BvP=vB+r˙(−cosθi+sinθj)+rθ˙(sinθi+cosθj)vP=(vB−r˙cosθ+rθ˙sinθ)i+(r˙sinθ+rθ˙cosθ)j$

We can rewrite this as,

$vP=vPxi+vPyj$

Where,

$vPx=vB−r˙cosθ+rθ˙sinθvPy=r˙sinθ+rθ˙cosθ$

To find the magnitude,

$vP=vPx2+vPy2$

Now, plot the graph,

Vector Mechanics For Engineers, Chapter 11.5, Problem 11.164P , additional homework tip 4

Conclusion:

The magnitude of velocity of the parasailer is equal to,

$vP=vPx2+vPy2$

The relevant graph is shown above.

To determine

(b)

The magnitude of acceleration of parasailer at $t=5 s$

Expert Solution

Answer to Problem 11.164P

$aP=1.7871 m/s2$

Explanation of Solution

Given information:

Vector Mechanics For Engineers, Chapter 11.5, Problem 11.164P , additional homework tip 5

Length of rope is defined as,

$r=600−18t5/2$

Constant velocity of the boat is $15 kn$

The angle is increasing at $2°s$

The velocity in radial and transverse components,

$v=vrer+vθeθ=r˙er+rθ˙eθ$

The acceleration in radial and transverse components,

$a=arer+aθeθ=(r¨−rθ˙2)er+(rθ¨+2r˙θ˙)eθ$

Calculation:

Vector Mechanics For Engineers, Chapter 11.5, Problem 11.164P , additional homework tip 6

According to given information,

Convert,

$vB=15 kn=25.3 ft/saB=0$

We know that,

$r=600−18t5/2$

$θ˙=2°s =0.0349 rad/s$

Therefore,

$r˙=−516t3/2r¨=−1532t1/2$

$θ¨=0$

According to the explanation, the relative acceleration of parasailer with respect to boat is,

$aP/B=(r¨−rθ˙2)er+(rθ¨+2r˙θ˙)eθaP/B=[−1532t1/2−(600−18t5/2)(0.0349 rad/s)2]er+[2(−516t3/2)(0.0349 rad/s)]eθ$

The acceleration of parasailer is equal to,

$aP=aB+aP/BaP=0+[(−1532t1/2−(600−18t5/2)(0.0349 rad/s)2)er+(2(−516t3/2)(0.0349 rad/s))eθ]$

At, $t=5 s$

$aP=[(−1532(5)1/2−(600−18(5)5/2)(0.0349 rad/s)2)er+(2(−516(5)3/2)(0.0349 rad/s))eθ]aP=−1.7704er−0.2439eθ$

Therefore the magnitude is equal to,

$aP=(−1.7704)2+(−0.2439)2aP=1.7871 m/s2$

Conclusion:

The magnitude of acceleration of the parasailer is equal to,

$aP=1.7871 m/s2$

Answer 6

Chapter 11.5, Problem 11.164P

To determine

(a)

The magnitude of the velocity of the parasailer as a function of time.

Expert Solution

Answer to Problem 11.164P

Vector Mechanics For Engineers, Chapter 11.5, Problem 11.164P , additional homework tip 1

Explanation of Solution

Given information:

Vector Mechanics For Engineers, Chapter 11.5, Problem 11.164P , additional homework tip 2

Length of rope is defined as,

$r=600−18t5/2$

Constant velocity of the boat is $15 kn$

The angle is increasing at $2°s$

The velocity in radial and transverse components,

$v=vrer+vθeθ=r˙er+rθ˙eθ$

The acceleration in radial and transverse components,

$a=arer+aθeθ=(r¨−rθ˙2)er+(rθ¨+2r˙θ˙)eθ$

Calculation:

Vector Mechanics For Engineers, Chapter 11.5, Problem 11.164P , additional homework tip 3

According to given information,

Convert,

$vB=15 kn=25.3 ft/saB=0$

We know that,

$r=600−18t5/2$

$θ˙=2°s =0.0349 rad/s$

Therefore,

$r˙=−516t3/2r¨=−1532t1/2$

$θ¨=0$

According to the explanation, the relative velocity of parasailer with respect to boat is,

$vP/B=r˙er+rθ˙eθ$

The relative co-ordinates of the relative velocity is equal to,

$vP/B=r˙(−cosθi+sinθj)+rθ˙(sinθi+cosθj)$

Therefore the velocity of parasailer is equal to,

$vP=vB+vP/BvP=vB+r˙(−cosθi+sinθj)+rθ˙(sinθi+cosθj)vP=(vB−r˙cosθ+rθ˙sinθ)i+(r˙sinθ+rθ˙cosθ)j$

We can rewrite this as,

$vP=vPxi+vPyj$

Where,

$vPx=vB−r˙cosθ+rθ˙sinθvPy=r˙sinθ+rθ˙cosθ$

To find the magnitude,

$vP=vPx2+vPy2$

Now, plot the graph,

Vector Mechanics For Engineers, Chapter 11.5, Problem 11.164P , additional homework tip 4

Conclusion:

The magnitude of velocity of the parasailer is equal to,

$vP=vPx2+vPy2$

The relevant graph is shown above.

Answer 7

Chapter 11.5, Problem 11.164P

To determine

(a)

The magnitude of the velocity of the parasailer as a function of time.

Answer 8

Expert Solution

Answer to Problem 11.164P

Vector Mechanics For Engineers, Chapter 11.5, Problem 11.164P , additional homework tip 1

Answer 9

Expert Solution

Answer 10

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Given information:

Vector Mechanics For Engineers, Chapter 11.5, Problem 11.164P , additional homework tip 2

Length of rope is defined as,

$r=600−18t5/2$

Constant velocity of the boat is $15 kn$

The angle is increasing at $2°s$

The velocity in radial and transverse components,

$v=vrer+vθeθ=r˙er+rθ˙eθ$

The acceleration in radial and transverse components,

$a=arer+aθeθ=(r¨−rθ˙2)er+(rθ¨+2r˙θ˙)eθ$

Calculation:

Vector Mechanics For Engineers, Chapter 11.5, Problem 11.164P , additional homework tip 3

According to given information,

Convert,

$vB=15 kn=25.3 ft/saB=0$

We know that,

$r=600−18t5/2$

$θ˙=2°s =0.0349 rad/s$

Therefore,

$r˙=−516t3/2r¨=−1532t1/2$

$θ¨=0$

According to the explanation, the relative velocity of parasailer with respect to boat is,

$vP/B=r˙er+rθ˙eθ$

The relative co-ordinates of the relative velocity is equal to,

$vP/B=r˙(−cosθi+sinθj)+rθ˙(sinθi+cosθj)$

Therefore the velocity of parasailer is equal to,

$vP=vB+vP/BvP=vB+r˙(−cosθi+sinθj)+rθ˙(sinθi+cosθj)vP=(vB−r˙cosθ+rθ˙sinθ)i+(r˙sinθ+rθ˙cosθ)j$

We can rewrite this as,

$vP=vPxi+vPyj$

Where,

$vPx=vB−r˙cosθ+rθ˙sinθvPy=r˙sinθ+rθ˙cosθ$

To find the magnitude,

$vP=vPx2+vPy2$

Now, plot the graph,

Vector Mechanics For Engineers, Chapter 11.5, Problem 11.164P , additional homework tip 4

Conclusion:

The magnitude of velocity of the parasailer is equal to,

$vP=vPx2+vPy2$

The relevant graph is shown above.

Answer 13

To determine

(b)

The magnitude of acceleration of parasailer at $t=5 s$

Expert Solution

Answer to Problem 11.164P

$aP=1.7871 m/s2$

Explanation of Solution

Given information:

Vector Mechanics For Engineers, Chapter 11.5, Problem 11.164P , additional homework tip 5

Length of rope is defined as,

$r=600−18t5/2$

Constant velocity of the boat is $15 kn$

The angle is increasing at $2°s$

The velocity in radial and transverse components,

$v=vrer+vθeθ=r˙er+rθ˙eθ$

The acceleration in radial and transverse components,

$a=arer+aθeθ=(r¨−rθ˙2)er+(rθ¨+2r˙θ˙)eθ$

Calculation:

Vector Mechanics For Engineers, Chapter 11.5, Problem 11.164P , additional homework tip 6

According to given information,

Convert,

$vB=15 kn=25.3 ft/saB=0$

We know that,

$r=600−18t5/2$

$θ˙=2°s =0.0349 rad/s$

Therefore,

$r˙=−516t3/2r¨=−1532t1/2$

$θ¨=0$

According to the explanation, the relative acceleration of parasailer with respect to boat is,

$aP/B=(r¨−rθ˙2)er+(rθ¨+2r˙θ˙)eθaP/B=[−1532t1/2−(600−18t5/2)(0.0349 rad/s)2]er+[2(−516t3/2)(0.0349 rad/s)]eθ$

The acceleration of parasailer is equal to,

$aP=aB+aP/BaP=0+[(−1532t1/2−(600−18t5/2)(0.0349 rad/s)2)er+(2(−516t3/2)(0.0349 rad/s))eθ]$

At, $t=5 s$

$aP=[(−1532(5)1/2−(600−18(5)5/2)(0.0349 rad/s)2)er+(2(−516(5)3/2)(0.0349 rad/s))eθ]aP=−1.7704er−0.2439eθ$

Therefore the magnitude is equal to,

$aP=(−1.7704)2+(−0.2439)2aP=1.7871 m/s2$

Conclusion:

The magnitude of acceleration of the parasailer is equal to,

$aP=1.7871 m/s2$

Answer 14

To determine

(b)

The magnitude of acceleration of parasailer at $t=5 s$

Answer 15

Expert Solution

Answer to Problem 11.164P

$aP=1.7871 m/s2$

Answer 16

Expert Solution

Answer 17

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Given information:

Vector Mechanics For Engineers, Chapter 11.5, Problem 11.164P , additional homework tip 5

Length of rope is defined as,

$r=600−18t5/2$

Constant velocity of the boat is $15 kn$

The angle is increasing at $2°s$

The velocity in radial and transverse components,

$v=vrer+vθeθ=r˙er+rθ˙eθ$

The acceleration in radial and transverse components,

$a=arer+aθeθ=(r¨−rθ˙2)er+(rθ¨+2r˙θ˙)eθ$

Calculation:

Vector Mechanics For Engineers, Chapter 11.5, Problem 11.164P , additional homework tip 6

According to given information,

Convert,

$vB=15 kn=25.3 ft/saB=0$

We know that,

$r=600−18t5/2$

$θ˙=2°s =0.0349 rad/s$

Therefore,

$r˙=−516t3/2r¨=−1532t1/2$

$θ¨=0$

According to the explanation, the relative acceleration of parasailer with respect to boat is,

$aP/B=(r¨−rθ˙2)er+(rθ¨+2r˙θ˙)eθaP/B=[−1532t1/2−(600−18t5/2)(0.0349 rad/s)2]er+[2(−516t3/2)(0.0349 rad/s)]eθ$

The acceleration of parasailer is equal to,

$aP=aB+aP/BaP=0+[(−1532t1/2−(600−18t5/2)(0.0349 rad/s)2)er+(2(−516t3/2)(0.0349 rad/s))eθ]$

At, $t=5 s$

$aP=[(−1532(5)1/2−(600−18(5)5/2)(0.0349 rad/s)2)er+(2(−516(5)3/2)(0.0349 rad/s))eθ]aP=−1.7704er−0.2439eθ$

Therefore the magnitude is equal to,

$aP=(−1.7704)2+(−0.2439)2aP=1.7871 m/s2$

Conclusion:

The magnitude of acceleration of the parasailer is equal to,

$aP=1.7871 m/s2$

Answer 20

Ch. 11.1 - A bus travels the 100 miles between A and B at 50...Ch. 11.1 - Two cars A and B race each other down a straight...Ch. 11.1 - A snowboarder starts from rest at the top of a...Ch. 11.1 - The motion of a particle is defined by the...Ch. 11.1 - The vertical motion of mass A is defined by the...Ch. 11.1 - A loaded railroad car is rolling at a constant...Ch. 11.1 - A group of hikers uses a GPS while doing a 40-mile...Ch. 11.1 - The motion of a particle is defined by the...Ch. 11.1 - A girl operates a radio-controlled model ear in a...Ch. 11.1 - The motion of a particle is defined by the...

(a) The magnitude of the velocity of the parasailer as a function of time.

Concept explainers

Videos

(a)

The magnitude of the velocity of the parasailer as a function of time.

Answer to Problem 11.164P

Explanation of Solution

(b)

The magnitude of acceleration of parasailer at $t=5 s$

Answer to Problem 11.164P

Explanation of Solution

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Chapter 11 Solutions

(a) The magnitude of the velocity of the parasailer as a function of time.

Concept explainers

Videos

(a) The magnitude of the velocity of the parasailer as a function of time.

Answer to Problem 11.164P

Explanation of Solution

(b) The magnitude of acceleration of parasailer at t=5 s<m:math><m:mrow><m:mi>t</m:mi><m:mo>=</m:mo><m:mn>5</m:mn><m:mtext> </m:mtext><m:mi>s</m:mi></m:mrow></m:math>

Answer to Problem 11.164P

Explanation of Solution

Want to see more full solutions like this?

Chapter 11 Solutions

(a)

The magnitude of the velocity of the parasailer as a function of time.

(b)

The magnitude of acceleration of parasailer at $t=5 s$