Question

(21 pts.) Two people are on a row boat out in the middle of a calm lake. Initially the boat and the people in the boat are at rest. The people, who are good swimmers, are standing in the boat and decide to race to shore. Since they are roughly in the middle of the lake, they decide to swim in opposite directions from one another. The people simultaneously dive from opposite ends of the boat. Person A, who weighs 110.224lbs, dives leftward from the left end of the boat and has a horizontal speed of 3.000m/s the moment they lose contact with the boat. Person B, who weighs 154.315lbs, dives rightward from the right side of the boat and has a horizontal speed of 3.125m/s the moment they lose contact with the boat. Given the contact time of each person while diving to when they loss contact is 0.775s for Person A and 0.795s for person B. The boat, itself, has a mass of 200.Okg. (Remember that direction is important, so use positive and negative signs whenever determining, or using, position, displacement, velocity, acceleration, or force values) NOTE: Treat motion, force on boat due to A, force on boat due to B, net force on boat because of person A and B diving off boat, acceleration of boat due to A acceleration of boat due to B, and net acceleration of boat as only acting along the horizontal direction. Also, take leftward to be negative) Determine a) the force on the boat due to person A, Fbout A b) the acceleration on the boat due to person A, about c) the force on the boat due to person B, Frost, d) the acceleration on the boat due to person, about e) the net horizontal force on the boat because person A and person B applied the forces found in parts (a) and (c) the net horizontal acceleration of the boat because person A and person B caused the boat to experience accelerations found in parts (b) and (c). 8) the net displacement of the boat because Person A and Person B dove off the boat.

Answer 1

Solution

The following scheme helps us better understand the problem

1 Boat Lake

a) The force on the boat due to person A

Data

$m_{A}=110.224lbs=49.997kg$

$v_{A}=3.000m/s$

$t_{A}=0.775s$

The following diagram represents the situation for person A

V

Applying Newton's second law, we have

$\vec{F_{b,A}}=\frac{{F_{b,A}}}{dt}=\frac{\Delta{P_{b,A}} }{\Delta t}(-\hat{i})$ (1)

Then

$\Delta P{_{b,A}}=m_{A}({v_{f}}_{A}-{v_{o}})$ (2)

But, the velocity initial is zero, for A:

Then

$\Delta P{_{b,A}}=49.997kg\ast (3.000m/s)=149.991kgm/s$

$\vec{F_{b,A}}=-\frac{149.991kgm/s}{0.755s}=-198.664N\hat{i}$

b)

$\vec{F_{b,A}}=-m_{A}a\hat{i}$

Clearing a:

$\vec{a}=-\frac{F_{b,A}}{m_{A}}\hat{i}$

$\vec{a}=-\frac{198.664N}{49.997kg}\hat{i}=-3.974m/s^{2}\hat{i}$

c)

Data

$m_{B}=154.315lbs=69.996kg$

$v_{fB}=3.12m/s$

$t_{B}=0.795s$

의

Applying Newton's second law, we have:

$\vec{F_{b,B}}=\frac{{F_{b,B}}}{dt}=\frac{\Delta{P_{b,B}} }{\Delta t}(\hat{i})$ (3)

$\Delta P{_{b,B}}=m_{B}({v_{f}}_{B}-{v_{o}})$

$\Delta P{_{b,B}}=69.996kg\ast (3.125m/s)=218.738kgm/s$

Then

$\vec{F_{b,B}}=\frac{218.738kgm/s}{0.795s}=275.142N\hat{i}$

d)

$\vec{F_{b,B}}=m_{B}a\hat{i}$

Then

$\vec{a}=\frac{F_{b,B}}{m_{B}}\hat{i}$

$\vec{a}=\frac{275.142N}{69.996kg}\hat{i}=3.931m/s^{2}\hat{i}$