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

Answer to Problem 11.164P
Explanation of Solution
Given information:
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=˙rer+r˙θeθ
The acceleration in radial and transverse components,
a=arer+aθeθ=(¨r−r˙θ2)er+(r¨θ+2˙r˙θ)eθ
Calculation:
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=−516t3/2¨r=−1532t1/2
¨θ=0
According to the explanation, the relative velocity of parasailer with respect to boat is,
vP/B=˙rer+r˙θeθ
The relative co-ordinates of the relative velocity is equal to,
vP/B=˙r(−cosθi+sinθj)+r˙θ(sinθi+cosθj)
Therefore the velocity of parasailer is equal to,
vP=vB+vP/BvP=vB+˙r(−cosθi+sinθj)+r˙θ(sinθi+cosθj)vP=(vB−˙rcosθ+r˙θsinθ)i+(˙rsinθ+r˙θcosθ)j
We can rewrite this as,
vP=vPxi+vPyj
Where,
vPx=vB−˙rcosθ+r˙θsinθvPy=˙rsinθ+r˙θcosθ
To find the magnitude,
vP=√vPx2+vPy2
Now, plot the graph,
Conclusion:
The magnitude of velocity of the parasailer is equal to,
vP=√vPx2+vPy2
The relevant graph is shown above.
(b)
The magnitude of acceleration of parasailer at t=5 s

Answer to Problem 11.164P
aP=1.7871 m/s2
Explanation of Solution
Given information:
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=˙rer+r˙θeθ
The acceleration in radial and transverse components,
a=arer+aθeθ=(¨r−r˙θ2)er+(r¨θ+2˙r˙θ)eθ
Calculation:
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=−516t3/2¨r=−1532t1/2
¨θ=0
According to the explanation, the relative acceleration of parasailer with respect to boat is,
aP/B=(¨r−r˙θ2)er+(r¨θ+2˙r˙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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Chapter 11 Solutions
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