Practice Midterm 1 Problem 2

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Practice Midterm 1 [ Edit ]

Overview Summary View Diagnostics View Print View with Answers 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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2/17/2018 ANSWER:

\(\texttip{v_{\rm max}}{v_max}\) = 56 \(\rm m/s\) Also accepted: 55, 57

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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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2/17/2018 ANSWER: \(\texttip{d_{\rm 1,2}}{d_1,2}\) = 40 \(\rm m\)

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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:

\(v_0\) =

= 13.2

Also accepted:

= 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 ,

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= 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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