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Practice Midterm 1
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Practice Midterm 1
Due: 5:00pm on Monday, February 19, 2018
To understand how points are awarded, read the Grading Policy for this assignment.
Problem 2.60
Description: A subway train starts from rest at a station and accelerates at a rate of 1.60 m/s^2 for 14.0 s . It runs at constant speed for 70.0 s and slows down
at a rate of 3.50 m/s^2 until it stops at the next station. (a) Find the total distance covered.
A subway train starts from rest at a station and accelerates at a rate of \(1.60\;{\rm m}/{\rm s}^{2}\) for 14.0 \({\rm s}\) . It runs at constant speed for 70.0 \({\rm s}\) and
slows down at a rate of \(3.50\;{\rm m}/{\rm s}^{2}\) until it stops at the next station.
Part A
Find the total distance covered.
ANSWER:
\(d\) = 1.80
\({\rm km}\)
Problem 1.79
Description: Vectors A_vec and B_vec have scalar product scalar and their vector product has magnitude vector. (a) What is the angle between these two
vectors?
Vectors \(\vec A\) and \(\vec B\) have scalar product -2.00 and their vector product has magnitude 5.00.
Part A
What is the angle between these two vectors?
ANSWER:
\(\texttip{\theta }{theta}\) =
= 112
\(^\circ\)
± The Graph of a Sports Car's Velocity
Description: ± Includes Math Remediation. Find an object's acceleration and distance traveled from a graph of velocity as a function of time.
The graph in the figure shows the velocity \(\texttip{v}{v}\) of a sports car as a function of time \(\texttip{t}{t}\). Use the
graph to answer the following questions.
Part A
Find the maximum velocity \(\texttip{v_{\rm max}}{v_max}\) of the car during the ten-second interval depicted in the graph.
Express your answer in meters per second to the nearest integer.
Hint 1. How to approach the problem
Because the graph displays the car's velocity at each moment in time, the maximum velocity of the car can be found simply by locating the maximum
value of the velocity on the graph.
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ANSWER:
\(\texttip{v_{\rm max}}{v_max}\) = 56
\(\rm m/s\)
Also accepted: 55, 57
Part B
During which time interval is the acceleration positive?
Indicate the best answer.
Hint 1. Finding acceleration from the graph
Recall that acceleration is the rate of change of velocity with respect to time. Therefore, on this graph of velocity vs. time, acceleration is the slope of the
graph. Recall that the slope \(\texttip{m}{m}\) is defined by \(m=\Delta y/\Delta x\) for a graph of \(\texttip{y}{y}\) vs. \(\texttip{x}{x}\), or \(m=\Delta v/\Delta
t\) in this case. If the graph is increasing from left to right, then the slope is positive.
ANSWER:
\(t=0\;{\rm s}\) to \(t=6\;{\rm s}\)
\(t=0\;{\rm s}\) to \(t=4\;{\rm s}\)
\(t=0\;{\rm s}\) to \(t=10\;{\rm s}\)
\(t=4\;{\rm s}\) to \(t=10\;{\rm s}\)
\(t=2\;{\rm s}\) to \(t=6\;{\rm s}\)
Part C
Find the maximum acceleration \(\texttip{a_{\rm max}}{a_max}\) of the car.
Express your answer in meters per second per second to the nearest integer.
Hint 1. How to approach the problem
The car's acceleration is the rate of change of the car's velocity \(\texttip{v}{v}\) with respect to time \(\texttip{t}{t}\). In this problem, the car's velocity is
given graphically, so the car's acceleration at a given moment is found from the slope of the \(\texttip{v}{v}\) vs. \(\texttip{t}{t}\) curve at that moment. If the
\(\texttip{v}{v}\) vs. \(\texttip{t}{t}\) curve over some time interval is represented by a straight line, the instantaneous acceleration anywhere in that interval
is equal to the slope of the line, that is, to the average acceleration over that time interval.
To find the maximum acceleration, find the value of the curve's greatest positive slope.
Hint 2. Find the final velocity on the interval with greatest acceleration
The slope of the curve is greatest during the first second of motion. The slope of the graph on this interval is given by the change in velocity divided by the
change in time over the interval from \(t=0\) to \(t=1\). At time \(t = 0\; {\rm s}\), the car's velocity \(v(0)\) is zero. Find the velocity \(\texttip{v\left(1\right)}
{v(1)}\) of the car at time \(t=1\;{\rm s}\).
Express your answer in meters per second to the nearest integer.
ANSWER:
\(\texttip{v\left(1\right)}{v(1)}\) = 30
\(\rm m/s\)
ANSWER:
\(\texttip{a_{\rm max}}{a_max}\) = 30
\(\rm m/s^2\)
Part D
Find the minimum magnitude of the acceleration \(\texttip{a_{\rm min}}{a_min}\) of the car.
Express your answer in meters per second per second to the nearest integer.
Hint 1. How to approach the problem
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To find the minimum magnitude of the acceleration of the car, you must find the point where the absolute value of the slope is smallest.
ANSWER:
\(\texttip{a_{\rm min}}{a_min}\) = 0
\(\rm m/s^2\)
Part E
Find the distance \(\texttip{d_{\rm 0,2}}{d_0,2}\) traveled by the car between \(t=0\;\rm s\) and \(t=2\; \rm s\).
Express your answer in meters to the nearest integer.
Hint 1. How to approach the problem
In this problem, the car's velocity as a function of time is given graphically, so the distance traveled is represented by the area under the \(\texttip{v}{v}\)
vs. \(\texttip{t}{t}\) graph between \(t=0\;\rm s\) and \(t=2 \; \rm s\).
Hint 2. Find the distance traveled in the first second
What is the distance \(\texttip{d_{\rm 0,1}}{d_0,1}\) traveled between \(t=0\;\rm s\) and \(t=1\;\rm s\)?
Express your answer in meters.
Hint 1. The area of a triangle
Observe that the region in question is a triangle , whose area is therefore one-half the
product of the base and the height.
ANSWER:
d_0,1 = 15
\(\rm m\)
Hint 3. Find the distance traveled in the second second
What is the distance \(\texttip{d_{\rm 1,2}}{d_1,2}\) traveled between \(t=1\;\rm s\) and\(t=2\;\rm s\)?
Express your answer in meters.
Hint 1. The shape of the region
The region under the graph between 1 and 2 seconds can be seen as consisting of a rectangle and a triangle.
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ANSWER:
\(\texttip{d_{\rm 1,2}}{d_1,2}\) = 40
\(\rm m\)
ANSWER:
\(\texttip{d}{d}\) = 55
\(\rm m\)
Exercise 2.52
Description: The acceleration of a bus is given by a_x(t)= alpha t, where alpha is a constant. (a) If the bus's velocity at time t_1 is v_1, what is its velocity at time
t_2? (b) If the bus's position at time t_1 is x_1, what is its position at time t_2?
The acceleration of a bus is given by \(a_{x}(t)= \alpha t\), where \(\texttip{\alpha}{alpha}\) = 1.12 \({\rm m/s^3}\) is a constant.
Part A
If the bus's velocity at time \(\texttip{t_1}{t_1}\) = 1.11 \({\rm s}\) is 4.94 \({\rm m/s}\) , what is its velocity at time \(\texttip{t_2}{t_2}\) = 2.07 \({\rm s}\) ?
ANSWER:
\(v\) =
= 6.65
\({\rm m/s}\)
Part B
If the bus's position at time \(\texttip{t_1}{t_1}\) = 1.11 \({\rm s}\) is 5.98 \({\rm m}\) , what is its position at time \(\texttip{t_2}{t_2}\) = 2.07 \({\rm s}\) ?
ANSWER:
\(x\) =
= 11.5
\({\rm m}\)
Problem 3.69
Description: In the middle of the night you are standing a horizontal distance of 14.0 m from the high fence that surrounds the estate of your rich uncle. The top
of the fence is 5.00 m above the ground. You have taped an important message to a rock that...
In the middle of the night you are standing a horizontal distance of 14.0 \(\rm m\) from the high fence that surrounds the estate of your rich uncle. The top of the fence
is 5.00 \(\rm m\) above the ground. You have taped an important message to a rock that you want to throw over the fence. The ground is level, and the width of the
fence is small enough to be ignored. You throw the rock from a height of 1.60 \(\rm m\) above the ground and at an angle of 54.0 \({\rm ^\circ}\) above the horizontal.
Part A
What minimum initial speed must the rock have as it leaves your hand to clear the top of the fence?
Express your answer with the appropriate units.
ANSWER:
\(v_0\) =
Also accepted:
= 13.2
= 13.2
,
= 13.2
Part B
For the initial velocity calculated in the previous part, what horizontal distance beyond the fence will the rock land on the ground?
Express your answer with the appropriate units.
ANSWER:
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\
(d\)
=
= 4.09
Also accepted:
= 4.09
,
= 4.09
Exercise 3.29
Description: At its Ames Research Center, NASA uses its large “20-G” centrifuge to test the effects of very large accelerations (“hypergravity”) on test pilots and
astronauts. In this device, an arm 8.84 m long rotates about one end in a horizontal plane, and the...
At its Ames Research Center, NASA uses its large “20-G” centrifuge to test the effects of very large accelerations (“hypergravity”) on test pilots and astronauts. In this
device, an arm 8.84 \({\rm m}\) long rotates about one end in a horizontal plane, and the astronaut is strapped in at the other end. Suppose that he is aligned along
the arm with his head at the outermost end. The maximum sustained acceleration to which humans are subjected in this machine is typically 12.5 \({\it g}\).
Part A
How fast must the astronaut's head be moving to experience this maximum acceleration?
ANSWER:
\(v\) = 32.9 \({\rm m/s}\)
Part B
What is the difference between the acceleration of his head and feet if the astronaut is 2.00 \({\rm m}\) tall?
ANSWER:
\(\Delta a\) = 27.7
\({\rm m/s^2}\)
Part C
How fast in rpm \(\left( {\rm rev/min} \right)\) is the arm turning to produce the maximum sustained acceleration?
ANSWER:
\(\large{\frac{1}{T}}\) = 35.5 \({\rm rpm}\)
Direction of Acceleration of Pendulum
Description: Questions about the direction and relative magnitude of acceleration at various points in the pendulum trajectory. Conceptual.
Learning Goal:
To understand that the direction of acceleration is in the direction of the change of the velocity, which is unrelated to the direction of the velocity.
The pendulum shown makes a full swing from \(-\pi/4\) to \(+ \pi/4\). Ignore friction and assume that the string is massless. The eight labeled arrows represent
directions to be referred to when answering the following questions.
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Part A
Which of the following is a true statement about the acceleration of the pendulum bob, \(\texttip{\vec{a}}{a_vec}\).
ANSWER:
\(\texttip{\vec{a}}{a_vec}\) is equal to the acceleration due to gravity.
\(\texttip{\vec{a}}{a_vec}\) is equal to the instantaneous rate of change in velocity.
\(\texttip{\vec{a}}{a_vec}\) is perpendicular to the bob's trajectory.
\(\texttip{\vec{a}}{a_vec}\) is tangent to the bob's trajectory.
Part B
What is the direction of \(\texttip{\vec{a}}{a_vec}\) when the pendulum is at position 1?
Enter the letter of the arrow parallel to \(\texttip{\vec{a}}{a_vec}\).
Hint 1. Velocity at position 1
What is the velocity of the bob when it is exactly at position 1?
ANSWER:
\(\texttip{v_{\rm 1}}{v_1}\) = 0
\({\rm m/s}\)
Hint 2. Velocity of bob after it has descended
What is the velocity of the bob just after it has descended from position 1?
ANSWER:
very small and having a direction best approximated by arrow D
very small and having a direction best approximated by arrow A
very small and having a direction best approximated by arrow H
The velocity cannot be determined without more information.
ANSWER:
H
Part C
What is the direction of \(\texttip{\vec{a}}{a_vec}\) at the moment the pendulum passes position 2?
Enter the letter of the arrow that best approximates the direction of \(\texttip{\vec{a}}{a_vec}\).
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Hint 1. Instantaneous motion
At position 2, the instantaneous motion of the pendulum can be approximated as uniform circular motion. What is the direction of acceleration for an
object executing uniform circular motion?
ANSWER:
C
We know that for the object to be traveling in a circle, some component of its acceleration must be pointing radially inward.
Part D
What is the direction of \(\texttip{\vec{a}}{a_vec}\) when the pendulum reaches position 3?
Give the letter of the arrow that best approximates the direction of \(\texttip{\vec{a}}{a_vec}\).
Hint 1. Velocity just before position 3
What is the velocity of the bob just before it reaches position 3?
ANSWER:
very small and having a direction best approximated by arrow B
very small and having a direction best approximated by arrow C
very small and having a direction best approximated by arrow H
The velocity cannot be determined without more information.
Hint 2. Velocity of bob at position 3
What is the velocity of the bob when it reaches position 3?
ANSWER:
\(\texttip{v_{\rm 3}}{v_3}\) = 0
\({\rm m/s}\)
ANSWER:
F
Part E
As the pendulum approaches or recedes from which position(s) is the acceleration vector \(\texttip{\vec{a}}{a_vec}\) almost parallel to the velocity vector \
(\texttip{\vec{v}}{v_vec}\).
ANSWER:
position 2 only
positions 1 and 2
positions 2 and 3
positions 1 and 3
Exercise 3.37
Description: Canada geese migrate essentially along a north-south direction for well over a thousand kilometers in some cases, traveling at speeds up to about
100 km/h. The one goose is flying at 100 km/h relative to the air but a ## -km/h wind is blowing ...
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Canada geese migrate essentially along a north-south direction for well over a thousand kilometers in some cases, traveling at speeds up to about 100 \({\rm km/h}\).
The one goose is flying at 100 \({\rm km/h}\) relative to the air but a 50 \({\rm \; km/h}\) -\(\rm km/h\) wind is blowing from west to east.
Part A
At what angle relative to the north-south direction should this bird head to travel directly southward relative to the ground?
Express your answer using three significant figures.
ANSWER:
\(\theta\) =
= 30.0
\(^\circ\) west of south
Part B
How long will it take the goose to cover a ground distance of 600 \({\rm \; km}\) from north to south? (Note: Even on cloudy nights, many birds can navigate using
the earth's magnetic field to fix the north-south direction.)
Express your answer using three significant figures.
ANSWER:
\(t\) =
= 6.93
\({\rm h}\)
± A Canoe on a River
Description: ± Includes Math Remediation. Find the magnitude and direction of the velocity of a canoe on a river, measured with respect to the river, given the
velocity of the canoe and the velocity of the current relative to the earth.
A canoe has a velocity of 0.440 \({\rm m/s}\) southeast relative to the earth. The canoe is on a river that is flowing at 0.570 \({\rm m/s}\) east relative to the earth.
Part A
Find the magnitude of the velocity \(\vec{v}_{\rm c/r}\) of the canoe relative to the river.
Express your answer in meters per second.
Hint 1. How to approach the problem
In this problem there are two reference frames: the earth and the river. An observer standing on the edge of the river sees the canoe moving at 0.440 \
({\rm m/s}\) , whereas an observer drifting with the river current perceives the canoe as moving with velocity \(\vec{v}_{\rm c/r}.\) Since the velocity of the
current in the river relative to the earth is known, you can determine \(\vec{v}_{\rm c/r}\) \(\). Note that the problem asks for the magnitude of \(\vec{v}_{\rm
c/r}\).\(\)
Hint 2. Find the relative velocity vector
Let \(\vec{v}_{\rm c/e}\) be the velocity of the canoe relative to the earth and \(\vec{v}_{\rm r/e}\) the velocity of the water in the river relative to the earth.
What is the velocity \(\vec{v}_{\rm c/r}\) of the canoe relative to the river?
Hint 1. Relative velocity
Consider a body A that moves with velocity \(\vec{v}_ {\rm A/S}\) relative to a reference frame S and with velocity \(\vec{v}_{\rm A/S'}\) relative to a
second reference frame \(\rm S'\). If \(S'\) moves with speed \(\vec{v}_{S'/S}\) relative to S, the velocity of the body relative to S is given by the
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vector sum
\(\vec{v}_{\rm A/S}=\vec{v}_{\rm A/S'}+\vec{v}_{\rm S'/S}\).
This equation is known as the Galilean transformation of velocity.
ANSWER:
\(\vec{v}_{\rm r/e}+\vec{v}_{\rm c/e}\)
\(\vec{v}_{\rm c/e}-\vec{v}_{\rm r/e}\)
\(\vec{v}_{\rm r/e}-\vec{v}_{\rm c/e}\)
Hint 3. Find the components of the velocity of the canoe relative to the river
Let the x axis point from west to east and the y axis from south to north. Find \((v_{ \rm c/r})_x\) and \((v_{\rm c/r})_y\), the x and the y components of the
velocity of the canoe relative to the river.
Express the two velocity components, separated by a comma, in meters per second.
Hint 1. How to approach the problem
The Galilean transformation of velocity tells you that the velocity of the canoe relative to the river is given by the difference of two vectors.
Therefore, the components of the velocity of the canoe relative to the river are given by the difference of the components of those two vectors.
Look back at the diagram from the introduction for help in setting up the equations.
Hint 2. Components of a vector
Consider a vector \(\texttip{\vec{A}}{A_vec}\) that forms an angle \(\texttip{\theta }{theta}\) with the positive x axis. The x and y components of \
(\texttip{\vec{A}}{A_vec}\) are
\(A_x=A\cos\theta\) and \(A_y=A\sin\theta,\)
where \(\texttip{A}{A}\) is the magnitude of the vector.
ANSWER:
,
\((v_{ \rm c/r})_x\) , \((v_{\rm c/r})_y\) =
= -0.259, -0.311
\(\rm m/s\)
Now simply calculate the magnitude of \(\vec{v}_{\rm c/r}\), which is given by the square root of the sum of the squares of its components.
ANSWER:
\(v_{\rm c/r}\) =
= 0.405
\(\rm m/s\)
Part B
Find the direction of the velocity of the canoe relative to the river.
Express your answer as an angle measured south of west.
Hint 1. How to approach the problem
The direction of a vector can be determined through simple trigonometric relations. You can use either the relation between the magnitude of the vector
and one of its components or the relation between the two components of the vector. In both cases, use the information found in Part A. Note that the
problem asks for the direction of \(\vec{v}_{\rm c/r}\) \(\) \(\) as an angle measured south of west; your answer should be a positive angle between \
(0^\circ\) and \(90^\circ\).
Hint 2. Find the direction of a vector given its components
Consider a vector of magnitude \(\texttip{A}{A}\) whose x component is \(\texttip{A_{\mit x}}{A_x}\) and y component is \(\texttip{A_{\mit y}}{A_y}\). What is
the angle this vector makes with the x axis?
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Hint 1. The direction of a vector
Consider a vector \(\texttip{\vec{A}}{A_vec}\) that forms an angle \(\texttip{\theta }{theta}\) with the positive x axis. The vector's x and y components
are
\(A_x=A\cos\theta\) and \(A_y=A\sin\theta,\)
where \(\texttip{A}{A}\) is the magnitude of the vector. Thus,
\(\large{\tan\theta=\frac{A_y}{A_x}}\), \(\large{\sin\theta = \frac{A_y}{A}}\), and \(\large{\cos\theta = \frac{A_x}{A}}\).
ANSWER:
\(\arcsin(A_y/A)\)
\(\arccos(A_y/A)\)
\(\arcsin(A/A_y)\)
\(\arctan(A_x/A_y)\)
\(\sin(A_y/A)\)
ANSWER:
= 50.2 degrees south of west
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