course Phy 201
002. Velocity
Question: `q001. Note that there are 14 questions in this assignment.
If an object moves 12 meters in 4 seconds, then at what average rate is the object moving? Explain how you obtained your result in terms of commonsense images.
Your solution: 3 m/s. I found this by dividing the 12 m by the 4 seconds.
Given Solution:
Moving 12 meters in 4 seconds, we move an average of 3 meters every second.
We can imagine dividing up the 12 meters into four equal parts, one for each second. Each part will span 3 meters, corresponding to the distance moved in 1 second, on the average.
Self-critique (if necessary):
Question: `q002. How is the preceding problem related to the concept of a rate?
Your solution: The concept of rate tells at what interval something is changing based on another. It is a simple ratio. The previous problem addresses rate in the sense that we are comparing the ratio of the quantity of meters to the quantity of seconds. This is found by dividing the change in the quantity of meters by the change in the quantity of seconds.
Given Solution:
A rate is obtained by dividing the change in a quantity by the change in another quantity on which is dependent. In this case we divided the change in position by the time during which that change occurred.
More specifically
The rate of change of A with respect to B is defined to be the quantity (change in A) / (change in B).
An object which moves 12 meters in 3 seconds changes its position by 12 meter during a change in clock time of 3 seconds. So the question implies
Change in position = 12 meters
Change in clock time = 3 seconds
When we divide the 12 meters by the 3 seconds we are therefore dividing (change in position) by (change in clock time). In terms of the definition of rate of change:
the change in position is the change in A, so position is the A quantity.
the change in clock time is the change in B, so clock time is the B quantity.
So
(12 meters) / (3 seconds) is
(change in position) / (change in clock time) which is the same as
average rate of change of position with respect to clock time.
Thus
average velocity is average rate of change of position with respect to clock time.
Self-critique (if necessary):
Question: `q003. Is object position dependent on time or is time dependent on object position?
Your solution: Object position is dependent on time. You divide the rate of change in meters by rate of change in time which it is dependent upon. No matter if the object is moving or not, time is always going, therefore position depends on time.
Given Solution:
Object position is dependent on time--the clock runs whether the object is moving or not so time is independent of position. Clock time is pretty much independent of anything else (this might not be so at the most fundamental level, but for the moment, unless you have good reason to do otherwise, this should be your convention).
Self-critique (if necessary)
Question: `q004. So the rate here is the average rate at which position is changing with respect to clock time. Explain what concepts, if any, you missed in your explanations.
Your solution: I believe I covered all concepts in my solutions.
Given Solution:
Be sure you have reviewed all the definitions and concepts associated with velocity. If there’s anything you don’t understand, be sure to address it in your self-critique.
Self-critique (if necessary):
Question: `q005. If an object is displaced -6 meters in three seconds, then what is the average speed of the object what is its average velocity? Explain how you obtained your result in terms of commonsense images and ideas.
Your solution: The average speed cannot be given because speed cannot be negative. The average velocity, however, is -2 m/s because it can be either positive or negative. You can think of this in terms of a person running a 100m dash. A person cannot be running at a negative speed, however his/her velocity, can be negative. If a person is observed beginning the race at the 100m mark and then 5 seconds into the race at the 50 m mark then he/she would have a displacement of -50 meters. His/her speed could not be negative because the time is going forward, but his/her position (ave velocity) could be. This is the same principle when calculating the average velocity of the object which was displaced -6 m in 3 seconds.
Given Solution:
Speed is the average rate at which distance changes with respect to clock time. Distance cannot be negative and the clock runs forward. Therefore speed cannot be negative.
Velocity is the average rate at which position changes with respect to clock time, and since position changes can be positive or negative, so can velocity.
In general distance has no direction, while velocity does have direction.
Putting it loosely, position is just how fast something is moving; velocity is how fast and in what direction.
Self-critique (if necessary):
Question: `q006. If `ds stands for the change in the position of an object and `dt for the time interval during which the position changes, then what expression stands for the average velocity vAve of the object during this time interval?
Your solution: vAve = `ds/ `dt
Given Solution:
Average velocity is rate of change of position with respect to clock time.
Change in position is `ds and change in clock time is `dt, so average velocity is expressed in symbols as
vAve = `ds / `dt.
Self-critique (if necessary)
Question: `q007. How do you write the expressions `ds and `dt on your paper?
Your solution: When written out, you use the symbol Delta (looks like a triangle) beside an x or an s, to indicate the displacement. Also use Delta beside a t to indicate time.
Given Solution:
You use the Greek capital Delta when writing on paper or when communicating outside the context of this course; this is the symbol that looks like a triangle. See Introductory Problem Set 1.
`d is used for typewritten communication because the symbol for Delta is not interpreted correctly by some Internet forms and text editors. You should get in the habit of thinking and writing the Delta symbol when you see `d.
You may use either `d or Delta when submitting work and answering questions.
Self-critique (if necessary):
Question: `q008. If an object changes position at an average rate of 5 meters/second for 10 seconds, then how far does it move?
How is this problem related to the concept of a rate?
Your solution: The object will move 50 meters. This is related to rate because the problem already states the rate at which the object is moving. It means that the change in position is 5 meters per second which is the time interval in which it is dependent upon. Then we are also given the change in time with the 10 seconds and therefore we can calculate how far the object moved by simply multiply the change in the object’s position in m/s and then multiply that by the 10 seconds.
Given Solution:
In this problem you are given the rate at which position changes with respect to time, and you are given the time interval during which to calculate the change in position. Given the rate at which quantity A changes with respect to quantity B, and the change in quantity B, we have
The given rate, which is `dA / `dB
The change in B, which is `dB
We can therefore multiply `dA / `dB by `dB to get
`dA / `dB * `dB = `dA.
That is, you can get the change in the first quantity (the A quantity) if you know the average rate (`dA / `dB) and the change in the second quantity (the B quantity).
Self-critique (if necessary)
Question: `q009. If vAve stands for the rate at which the position of the object changes with respect to clock time (also called velocity) and `dt for the time interval during which the change in position is to be calculated, then how do we write the expression for the change `ds in the position?
Your solution: `ds = vAve * `dt or Delta s = vAve * Delta t
Given Solution:
To find the change in a quantity we multiply the rate by the time interval during which the change occurs.
The velocity is the rate, so we obtain the change in position by multiplying the velocity by the time interval:
`ds = vAve * `dt.
The units of this calculation pretty much tell us what to do:
We know what it means to multiply pay rate by time interval (dollar / hr * hours of work) or automobile velocity by the time interval (miles / hour * hour).
When we multiply vAve, for example in in units of cm / sec or meters / sec, by `dt in seconds, we get displacement in cm or meters. Similar reasoning applies if we use different measures of distance.
Self-critique (if necessary):
Question: `q010. Explain how the quantities average velocity vAve, time interval `dt and displacement `ds are related by the definition of a rate, and how this relationship can be used to solve the current problem.
Your solution: The definition of rate is the change in quantity A divided by the change in quantity B. The solution you would get in terms of the above quantities would be the average velocity. The change in quantity A is the displacement and the change in quantity B in which it is being divided by is the time interval. Thus this gives you vAve= `ds / `dt.
Given Solution:
vAve is the average rate at which position changes. The change in position is the displacement `ds, the change in clock time is `dt, so vAve = `ds / `dt.
Self-critique (if necessary):
Question: `q011. The basic rate relationship vAve = `ds / `dt expresses the definition of average velocity vAve as the rate at which position s changes with respect to clock time t. What algebraic steps do we use to solve this equation for `ds, and what is our result?
Your solution: To solve for `ds you must multiply each side of the equation vAve = `ds / `dt by `dt. So, you would have `dt * vAVE = `ds / `dt * `dt which would ultimately leave you with only `ds on one side of the equals sign, because it cancels out the `dt on that side and multiplies it to the vAve, thus solving for `ds.
Given Solution:
To solve vAve = `ds / `dt for `ds, we multiply both sides by `dt. The steps:
vAve = `ds / `dt. Multiply both sides by `dt:
vAve * `dt = `ds / `dt * `dt Since `dt / `dt = 1
vAve * `dt = `ds . Switching sides we have
`ds = vAve * `dt.
Self-critique (if necessary):
Question: `q012. How is the preceding result related to our intuition about the meanings of the terms average velocity, displacement and clock time?
Your solution:
When you multiply your vAve by the length of time `dt, you end up with your original displacement `ds, which is the change of position during that length of time.
Given Solution:
For most of us our most direct intuition about velocity probably comes from watching an automobile speedometer.
We know that if we multiply our average velocity in mph by the duration `dt of the time interval during which we travel, we get the distance traveled in miles. From this we easily extend the idea.
Whenever we multiply our average velocity by the duration of the time interval, we expect to obtain the displacement, or change in position, during that time interval.
Self-critique (if necessary):
Question: `q013. What algebraic steps do we use to solve the equation vAve = `ds / `dt for `dt, and what is our result?
Your solution: You begin by multiplying both sides by `dt. That gives you `dt * vAve = `ds / `dt * `dt. When you do this it eliminates the `dt on the right side of the =, because `dt / `d gives you 1. Then all you have to do is divide both sides by vAve, which leaves you with `dt = `ds * vAve.
Given Solution:
To solve vAve = `ds / `dt for `dt, we must get `dt out of the denominator. Thus we first multiply both sides by the denominator `dt. Then we can see where we are and takes the appropriate next that. The steps:
vAve = `ds / `dt. Multiply both sides by `dt:
vAve * `dt = `ds / `dt * `dt Since `dt / `dt = 1
vAve * `dt = `ds. We can now divide both sides by vAve to get `dt = `ds / vAve.
Self-critique (if necessary):
Question: `q014. How is this result related to our intuition about the meanings of the terms average velocity, displacement and clock time?
Your solution: In order to find the duration of the time interval, we have to use the concept of dividing the displacement by the velocity. This can be thought of in terms of taking a trip in a car and determining how long it will take based on mph and the distance we are traveling.
Given Solution:
If we want to know how long it will take to make a trip at a certain speed, we know to divide the distance in miles by the speed in mph.
If we divide the number of miles we need to travel by the number of miles we travel in hour, we get the number of hours required. This is equivalent to the calculation `dt = `ds / vAve.
We extend this to the general concept of dividing the displacement by the velocity to get the duration of the time interval.
When dealing with displacement, velocity and time interval, we can always check our thinking by making the analogy with a simple example involving miles, hours and miles/hour.