Assignment 1: Kinematics in 2 and 3 Dimensions

Due: 2:00am on Friday, February 8, 2013

Note: To understand how points are awarded, read your instructor's Grading Policy.

Exercise 2.14

The figure shows the velocity of a solar-poweredcar as a function of time. The driver acceleratesfrom a stop sign, cruises for 20 at a constant

speed of 60 , and then brakes to come to a

stop 40 after leaving the stop sign.

Part A

Compute the average acceleration during the time interval to .

Express your answer using two significant figures.

ANSWER:

Correct

Part B

Compute the average acceleration during the time interval to .

Express your answer using two significant figures.

ANSWER:

Correct

PC1431AY1213SEM2

Assignment 1: Kinematics in 2 and 3 D... Resources

= 1.7

= -1.7

Signed in as Mikael Lemanza Help Close

Part C

Compute the average acceleration during the time interval to .

Express your answer using two significant figures.

ANSWER:

Correct

Part D

Compute the average acceleration during the time interval to .

Express your answer using two significant figures.

ANSWER:

Correct

Part E

What is the instantaneous acceleration at ?

Express your answer using two significant figures.

ANSWER:

Correct

Part F

What is the instantaneous acceleration at ?

Express your answer using two significant figures.

ANSWER:

Correct

= 0

= 0

= 0

= -1.7

± Arrow Hits Apple

An arrow is shot at an angle of above the horizontal. The arrow hits a tree a horizontal distance

away, at the same height above the ground as it was shot. Use for the

magnitude of the acceleration due to gravity.

Part A

Find , the time that the arrow spends in the air.

Answer numerically in seconds, to two significant figures.

Hint 1. Find the initial upward component of velocity in terms of D.

Introduce the (unknown) variables and for the initial components of velocity. Then use

kinematics to relate them and solve for . What is the vertical component of the initial

velocity?

Express your answer symbolically in terms of and .

Hint 1. Find

Find the horizontal component of the initial velocity.

Express your answer symbolically in terms of and given symbolic quantities.

ANSWER:

Hint 2. Find

What is the vertical component of the initial velocity?

Express your answer symbolically in terms of .

ANSWER:

ANSWER:

Hint 2. Find the time of flight in terms of the initial vertical component of velocity.

From the change in the vertical component of velocity, you should be able to find in terms of

and .

=

=

=

Give your answer in terms of and .

Hint 1. Find

When applied to the y-component of velocity, in this problem the formula for with

constant acceleration is

What is , the vertical component of velocity when the arrow hits the tree?

Answer symbolically in terms of only.

ANSWER:

ANSWER:

Hint 3. Put the algebra together to find symbolically.

If you have an expression for the initial vertical velocity component in terms in terms of and , and another in terms of and , you should be able to eliminate this initial

component to find an expression for

Express your answer symbolically in terms of given variables.

ANSWER:

ANSWER:

Correct

Suppose someone drops an apple from a vertical distance of 6.0 meters, directly above the point where thearrow hits the tree.

Part B

How long after the arrow was shot should the apple be dropped, in order for the arrow to pierce the

=

=

=

= 6.7

apple as the arrow hits the tree?

Express your answer numerically in seconds, to two significant figures.

Hint 1. When should the apple be dropped

The apple should be dropped at the time equal to the total time it takes the arrow to reach thetree minus the time it takes the apple to fall 6.0 meters.

Hint 2. Find the time it takes for the apple to fall 6.0 meters

How long does it take an apple to fall 6.0 meters?

Express your answer numerically in seconds, to two significant figures.

ANSWER:

ANSWER:

Correct

Circular Launch

A ball is launched up a semicircular chute in such a way that at the top of the chute, just before it goes intofree fall, the ball has a centripetal acceleration of magnitude 2 .

Part A

How far from the bottom of the chute does the ball land?

Your answer for the distance the ball travels from the end of the chute should contain .

= 1.1

= 5.6

Hint 1. Speed of ball upon leaving chute

How fast is the ball moving at the top of the chute?

Hint 1. Equation of motion

The centripetal acceleration for a particle moving in a circle is , where is its

speed and is its instantaneous radius of rotation.

ANSWER:

Hint 2. Time of free fall

How long is the ball in free fall before it hits the ground?

Express the free-fall time in terms of and .

Hint 1. Equation of motion

There is constant acceleration due to gravity, so you can use the general expression

.

Write the values of , , and (separated by commas) that are appropriate for this

situation. Use the standard convention that is the magnitude of the acceleration due to

gravity. Take at the ground, and take the positive y direction to be upward.

ANSWER:

Hint 2. Equation for the height of the ball

To find the time in free fall before the ball hits the ground, , set the general equation for

the height equal to the height of the ground.

Answer in terms of , , and .

ANSWER:

ANSWER:

=

, , = , ,

=

Hint 3. Finding the horizontal distance

The horizontal distance follows from , where . and were

found in Parts i and ii respectively.

ANSWER:

Correct

Problem 3.50

Spiraling Up. It is common to see birds of prey rising upward on thermals. The paths they take may bespiral-like. You can model the spiral motion as uniform circular motion combined with a constant upwardvelocity. Assume a bird completes a circle of radius 8.00 m every 5.00 s and rises vertically at a rate of3.00 m/s.

Part A

Find the speed of the bird relative to the ground.

ANSWER:

Correct

Part B

Find the magnitude of the bird's acceleration.

ANSWER:

Correct

Part C

Find the direction of the bird's acceleration.

ANSWER:

=

=

10.5 m/s

12.6

Correct

Part D

Find the angle between the bird's velocity vector and the horizontal.

ANSWER:

Correct

Problem 3.69

Two tanks are engaged in a training exercise on level ground. The first tank fires a paint-filled training roundwith a muzzle speed of 245 at an angle 14.4 above the horizontal while advancing toward the second

tank with a speed of 13.5 relative to the ground. The second tank is retreating at a speed of 31.5

relative to the ground, but is hit by the shell. You can ignore air resistance and assume the shell hits at thesame height above ground from which it was fired.

Part A

Find the distance between the tanks when the round was first fired.

Take free fall acceleration to be = 9.80 .

ANSWER:

Correct

Part B

Find the distance between the tanks at the time of impact.

Take free fall acceleration to be = 9.80 .

ANSWER:

Correct

0.0 above the horizontal

16.6

2730 m

2950 m

Battleship Shells

A battleship simultaneously fires two shells toward two identical enemy ships. One shell hits ship A, whichis close by, and the other hits ship B, which is farther away. The two shells are fired at the same speed.Assume that air resistance is negligible and that the magnitude of the acceleration due to gravity is .

Note that after Part B the question setup changes slightly.

Part A

What shape is the trajectory (graph of y vs. x) of the shells?

ANSWER:

Correct

Part B

For two shells fired at the same speed which statement about the horizontal distance traveled iscorrect?

Hint 1. Two things to consider

The distance traveled is the product of the x component of the velocity and the time in the air.How does the y component of the velocity affect the "air time"? What angle would give thelongest range?

ANSWER:

Correct

Now, consider for the remaining parts of the question below that both shells are fired at an angle greaterthan 45 degrees with respect to the horizontal. Remember that enemy ship A is closer than enemy ship B.

Part C

Which shell is fired at the larger angle?

straight line

parabola

hyperbola

The shape cannot be determined.

The shell fired at a larger angle with respect to the horizontal lands farther away.

The shell fired at an angle closest to 45 degrees lands farther away.

The shell fired at a smaller angle with respect to the horizontal lands farther away.

The lighter shell lands farther away.

Hint 1. Consider the limiting case

Consider the case in which a shell is fired at 90 degrees above the horizontal (i.e., straight up).What distance will the shell travel? Now lower the angle at which the shell is fired. What

happens to the distance ?

ANSWER:

Correct

Part D

Which shell is launched with a greater vertical velocity, ?

ANSWER:

Correct

Part E

Which shell is launched with a greater horizontal velocity, ?

ANSWER:

Correct

Part F

Which shell reaches the greater maximum height?

Hint 1. What determines maximum height?

A

B

Both shells are fired at the same angle.

A

B

Both shells are launched with the same vertical velocity.

A

B

Both shells are launched with the same horizontal velocity.

What determines the maximum height reached by the shell?

ANSWER:

ANSWER:

Correct

Part G

Which shell has the longest travel time (time elapsed between being fired and hitting the enemy ship)?

Hint 1. Consider the limiting case

If a shell is fired exactly horizontally (0 degrees) the shell hits the ground right away. As theangle above the horizontal increases, what happens to the time of travel? Does this change asthe angle becomes greater than 45 degrees?

ANSWER:

Correct

A Wild Ride

A car in a roller coaster moves along a track that consists of a sequence of ups and downs. Let the x axisbe parallel to the ground and the positive y axis point upward. In the time interval from to s, the

trajectory of the car along a certain section of the track is given by

,

where is a positive dimensionless constant.

horizontal velocity

vertical velocity

mass of the shell

A

B

Both shells reach the same maximum height.

A

B

Both shells have the same travel time.

Part A

At is the roller coaster car ascending or descending?

Hint 1. How to approach the problem

The direction of motion of a particle is given by the direction of its velocity. In this particularcase, you need to establish if velocity of the car points upward or downward. That can be easilydetermined by simply looking at the sign of the vertical component of the velocity of the car.

Hint 2. Find the vertical component of the velocity of the car

Find the y component of the velocity of the car, , at .

Express your answer in meters per second in terms of .

Hint 1. Velocity components

Consider a particle moving in the xy plane with position vector . The instantaneous

velocity of the particle is defined as the time derivative of , that is,

.

Then, the components of the velocity vector are

.

ANSWER:

Correct

What does a negative y component of velocity tell you about the direction of the motion ofthe car?

ANSWER:

Correct

Part B

Derive a general expression for the speed of the car.

=

ascending

descending

Express your answer in meters per second in terms of and .

Hint 1. How to approach the problem

The speed of a particle is the magnitude of the velocity vector of the particle. Since themagnitude of a vector depends on its components, to find the speed of the car you need toknow the components of the car's velocity.

Hint 2. Magnitude of a vector

The magnitude of a vector , whose components are and , is given by

.

Hint 3. Find the components of the velocity of the car

Find a general expression for and , that is, the x and y components of the velocity of the

car.

Express your answers in meters per second in terms of and . Separate the velocity

components with a comma.

Hint 1. The velocity vector

Consider a particle moving in the xy plane with position vector . The instantaneous

velocity of the particle is defined as the time derivative of , that is,

.

Then, the components of the velocity vector are

.

ANSWER:

Correct

ANSWER:

Correct

, =

=

Part C

The roller coaster is designed according to safety regulations that prohibit the speed of the car fromexceeding . Find the maximum value of allowed by these regulations.

Express your answer using two significant figures.

Hint 1. How to approach the problem

To comply with the regulations, the speed of the car cannot exceed the given safety limit at anytime. Thus, you need to determine what the maximum value of the speed is and impose thecondition that such a value cannot be greater than the safety limit.

Hint 2. Find the maximum value of the speed

Given the expression found in Part B, find the maximum speed of the car in terms of .

Express your answer in meters per second.

Hint 1. Using the calculus

Recall that a function has a local maximum at if and

, where

and .

If you are trying to find the maximum value of a function over a fixed interval, you mustalso check whether the function is maximized at one of the endpoints of the interval.

It might help to sketch the velocity versus time.

Hint 2. Find the first derivative of the speed

As you found in Part B, the speed of the car is described by the following function:

.

Find an expression for the first derivative of the speed with respect to time.

Express your answer as a function of and .

ANSWER:

Correct

Note that the first derivative of the speed is not necessarily equal to the magnitudeof the car's acceleration. In two dimensions, to find the magnitude of theacceleration, you would first need to find the x and y components of theacceleration, then compute the magnitude of the acceleration vector.

=

Hint 3. Find the time at which the speed reaches its maximum value

At what time does the speed reach its maximum value?

Remember to check not only the points where the derivative of the function is zero butalso the endpoints of the interval to .

Express your answer in seconds.

ANSWER:

Correct

Now calculate the value of the speed of the car at .

ANSWER:

ANSWER:

Correct

Shooting over a Hill

A projectile is fired with speed at an angle from the horizontal as shown in the figure .

= 2

=

= 1.7

Part A

Find the highest point in the trajectory, .

Express the highest point in terms of the magnitude of the acceleration due to gravity , the

initial velocity , and the angle .

Hint 1. Velocity at the top

At the highest point of the trajectory , .

Hint 2. Which equation to use

The three kinematic equations that govern the motion in the y direction are

,

,

and

.

The third equation contains the height variable , and all the other quantities are known (at

), so you could use it to find . This would be the simplest method.

Alternately, if you prefer, you could first find the time required to reach the maximum height fromthe first equation, and then use this time in the second equation to solve for .

ANSWER:

Correct

Part B

What is the range of the projectile, ?

Express the range in terms of , , and .

Hint 1. Find the total time spent in air

Find the total time the projectile spends in the air, by considering the time it takes to reach

the highest point (found in Part A) and then the time it takes to fall back to the ground.

Express your answer in terms of , , and .

ANSWER:

=

Correct

Hint 2. Find

What is the x coordinate of the projectile's position?

Express your answer in terms of , , and .

ANSWER:

Correct

ANSWER:

Correct

Consider your advice to an artillery officer who has the following problem. From his current postition, hemust shoot over a hill of height at a target on the other side, which has the same elevation as his gun.

He knows from his accurate map both the bearing and the distance to the target and also that the hill is

halfway to the target. To shoot as accurately as possible, he wants the projectile to just barely pass abovethe hill.

Part C

Find the angle above the horizontal at which the projectile should be fired.

Express your answer in terms of and .

Hint 1. How to approach the problem

In the first half of this problem, you found and in terms of and . Solve these two

equations to find in terms of and .

Hint 2. Set up the ratio

Find the ratio of to .

The only variable in your answer should be .

=

=

=

ANSWER:

Correct

ANSWER:

Correct

Recall the following trigonometry formulas:

,

,

and

.

In this case, since , you

can draw a right triangle with as one of the angles, an "opposite" side of length , and an

"adjacent" side of length . You can then use this triangle to find and , after you

find the length of the hypotenuse using the Pythagorean Theorem.

Part D

What is the initial speed?

Express in terms of , , and .

=

=

Hint 1. How to approach this part

Use one of the equations that you had derived for and . You will need to find an expression

for and/or to find .

Hint 2. Find

Use the expression you derived for and the Pythagorean theorem to find .

Leave your answer in terms of and

ANSWER:

Correct

Hint 3. Find

Now find the expression for .

Leave your answer in terms of and .

ANSWER:

Correct

Now use one of the equations that you had derived for and and the expression for

and/or from the previous hints to find .

ANSWER:

Correct

Part E

Find , the flight time of the projectile.

=

=

=

Express the flight time in terms of and .

Hint 1. How to proceed

First, find in terms of and . You can use hints in Part B in the previous half.

ANSWER:

ANSWER:

Correct

Uniform Circular Motion

Learning Goal:

To find the velocity and acceleration vectors for uniform circular motion and to recognize that thisacceleration is the centripetal acceleration.

Suppose that a particle's position is given by the following expression:

.

Part A

Choose the answer that best completes the following sentence:The particle's motion at can be described by ____________.

ANSWER:

=

=

an ellipse starting at time on the positive x axis

an ellipse starting at time on the positive y axis

a circle starting at time on the positive x axis

a circle starting at time on the positive y axis

Correct

The quantity (Greek letter omega) is defined to be the angular velocity of the particle. Note that

must have units of radians per second. If is constant, the particle is said to undergo uniform

circular motion.

Part B

When does the particle first cross the negative x axis?

Express your answer in terms of some or all of the variables (Greek letter omega), , and

.

ANSWER:

Correct

Now, consider the velocity and speed of the particle.

Part C

Find the particle's velocity as a function of time.

Express your answer using unit vectors (e.g., + , where and are functions of ,

, , and ).

Hint 1. Derivative of

In the problem statement, you were given . The velocity of the particle will be

.

Find the derivative of with respect to time.

Express your answer in terms of some or all of the variables , , , and .

ANSWER:

Correct

ANSWER:

=

=

Correct

Part D

Find the speed of the particle at time .

Express your answer in terms of some or all of the variables , , and .

Hint 1. Definition of the magnitude of a vector

The magnitude of is . Use this to find the speed, which is just the magnitude

of the velocity.

Hint 2. Complete an mportant trig identity

Complete the following fundamental trigonometric identity:

ANSWER:

Correct

ANSWER:

Correct

Note that the speed of the particle is constant: .

Part E

Now find the acceleration of the particle.

Express your answer using unit vectors (e.g., + , where and are functions of ,

, , and ).

ANSWER:

=

= 1

=

=

Correct

Part F

Your calculation is actually a derivation of the centripetal acceleration. To see this, express theacceleration of the particle in terms of its position .

Express your answer in terms of some or all of the variables and .

ANSWER:

Correct

Part G

Now find the magnitude of the acceleration as a function of time.

Express your answer in terms of some or all of the variables , , and .

ANSWER:

Correct

Part H

Finally, express the magnitude of the particle's acceleration in terms of and using the expression

you obtained for the speed of the particle.

Express your answer in terms of one or both of the variables and .

ANSWER:

Correct

There are three important things to remember about centripetal acceleration:

1. The centripetal acceleration is simply the acceleration of a particle going around in acircle.

2. It has magnitude of either or .

3. It is directed radially inward.

=

=

=

Curved Motion Diagram

The motion diagram shown in the figure represents a pendulum released from rest at an angle of 45 from

the vertical. The dots in the motion diagramrepresent the positions of the pendulum bob ateleven moments separated by equal timeintervals. The green arrows represent the averagevelocity between adjacent dots. Also given is a"compass rose" in which directions are labeledwith the letters of the alphabet.

Part A

What is the direction of the acceleration of the object at moment 5?

Enter the letter of the arrow with this direction from the compass rose in the figure. Type Z ifthe acceleration vector has zero length.

Hint 1. How to approach the problem

The acceleration of the object at moment 5 is the acceleration found from the change in velocitybetween moments 4 and 5 and moments 5 and 6.

Hint 2. Definition of acceleration

Acceleration is defined as the change in velocity per unit time. Mathematically,

.

Since velocity is a vector, acceleration is a vector that points in the direction of the change inthe velocity.

Hint 3. Change of velocity: a graphical interpretation

Let us assume that in a second the velocity of an object changes from an initial value to a

final value . Then, the change in velocity that the object undergoes in that interval of time is

. If one represents the velocity of the object graphically with vectors, then the

change of velocity can be evaluated simply by applying the rule of subtraction of vectors, asshown in the picture.

ANSWER:

Correct

Part B

What is the direction of the acceleration of the object at moments 0 and 10?

Enter the letters corresponding to the arrows with these directions from the compass rose inthe figure, separated by commas. Type Z if the acceleration vector has zero length.

Hint 1. Find the direction of the velocity

What is the direction of the velocity of this object at moments 1 and 9?

Enter the letters of the corresponding directions from the compass rose, separated bycommas. Type Z if the velocity vector has zero length.

ANSWER:

Correct

Hint 2. Definition of acceleration

Acceleration is defined as the change in velocity per unit time. Mathematically,

.

Since velocity is a vector, acceleration is a vector that points in the direction of the change inthe velocity.

Hint 3. Applying the definition of acceleration

A

directions at time step 1, time step 9 = D,B

To find the acceleration at moment 0, subtract the (vector) velocity at moment 0 from thevelocity at moment 1. Similarly, to find the acceleration at moment 10, subtract the (vector)velocity at moment 9 from the velocity at moment 10.

ANSWER:

Correct

± A Canoe on a River

A canoe has a velocity of 0.360 southeast relative to the earth. The canoe is on a river that is flowing at

0.580 east relative to the earth.

Part A

Find the magnitude of the velocity 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 onthe edge of the river sees the canoe moving at 0.360 , whereas an observer drifting with the

river current perceives the canoe as moving with velocity Since the velocity of the current in

the river relative to the earth is known, you can determine . Note that the problem asks for

the magnitude of .

Hint 2. Find the relative velocity vector

Let be the velocity of the canoe relative to the earth and the velocity of the water in the

river relative to the earth. What is the velocity of the canoe relative to the river?

directions at time step 0, time step 10 = D,F

Hint 1. Relative velocity

Consider a body A that moves with velocity relative to a reference frame S and with

velocity relative to a second reference frame . If moves with speed

relative to S, the velocity of the body relative to S is given by the vector sum

.

This equation is known as the Galilean transformation of velocity.

ANSWER:

Correct

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 and

, 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 tothe river is given by the difference of two vectors. Therefore, the components of thevelocity of the canoe relative to the river are given by the difference of the components ofthose two vectors. Look back at the diagram from the introduction for help in setting upthe equations.

Hint 2. Components of a vector

Consider a vector that forms an angle with the positive x axis. The x and y

components of are

and

where is the magnitude of the vector.

ANSWER:

, = -0.325,-0.255

Correct

Now simply calculate the magnitude of , which is given by the square root of the sum of

the squares of its components.

ANSWER:

Correct

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 useeither the relation between the magnitude of the vector and one of its components or the relationbetween 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 as an angle measured south of west; your

answer should be a positive angle between and .

Hint 2. Find the direction of a vector given its components

Consider a vector of magnitude whose x component is and y component is . What is

the angle this vector makes with the x axis?

Hint 1. The direction of a vector

Consider a vector that forms an angle with the positive x axis. The vector's x and y

components are

and

where is the magnitude of the vector. Thus,

, , and .

ANSWER:

= 0.413

Correct

ANSWER:

Correct

Problem 3.34

The Ferris wheel in the figure , which rotatescounterclockwise, is just starting up. At a given instant, apassenger on the rim of the wheel and passing through thelowest point of his circular motion is moving at 3.00 m/s andis gaining speed at a rate of .

Part A

Find the magnitude of the passenger's acceleration at this instant.

ANSWER:

Correct

38.0 degrees south of west

0.814

Part B

Find the direction of the passenger's acceleration at this instant.

ANSWER:

Correct

Problem 3.78

You are flying in a light airplane spotting traffic for a radio station. Your flight carries you due east above ahighway. Landmarks below tell you that your speed is 47.0 relative to the ground and your air speed

indicator also reads 47.0 . However, the nose of your airplane is pointed somewhat south of east and

the station's weather person tells you that the wind is blowing with speed 20.0 .

Part A

In which direction is the wind blowing?

Express your answer as an angle measured east of north.

ANSWER:

Correct

Score Summary:

Your score on this assignment is 100%.You received 50 out of a possible total of 50 points.

37.9 to the right of vertical

12.3 east of north