course Phy 231 002. Velocity*********************************************
.............................................
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):OK Self-critique rating:3 ********************************************* Question: `q002. How is the preceding problem related to the concept of a rate? ********************************************* Your solution: In the problem, they ask for a rate. Velocity is the rate of change that they are looking for. A rate is a ratio of two different quantities. ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ Confidence Assessment: 3
.............................................
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): ok Self-critique rating: 3 ********************************************* Question: `q003. Is object position dependent on time or is time dependent on object position? ********************************************* Your solution: The position is dependent on time. Time is almost always the independent variable because it goes on no matter what. ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ Confidence Assessment: 3
.............................................
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): OK Self-critique rating: 3 ********************************************* 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 understand the concepts of rates of change and dependent and independent variables. ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ Confidence Assessment: 3
.............................................
Given Solution: Be sure you have reviewed all the definitions and concepts associated with velocity. If theres anything you dont understand, be sure to address it in your self-critique. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK Self-critique rating: 3 ********************************************* 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: Velocity = distance/time so the velocity would be -2 m/s. However, the speed would be 2m/s because speed cannot be negative. Velocity is the speed AND direction of an object, while speed is just the rate of change. ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ Confidence Assessment: 3
.............................................
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): OK Self-critique rating: 3 ********************************************* 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: If ds represents change in position and dt represents the change in time, then velocity would be ds/dt. ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ Confidence Assessment: 3
.............................................
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): OK Self-critique rating: 3 ********************************************* Question: `q007. How do you write the expressions `ds and `dt on your paper? ********************************************* Your solution: ds and dt would be written using the Greek symbol delta. It resembles a triangle. ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ Confidence Assessment:
.............................................
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): OK Self-critique rating:3 ********************************************* 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: It would travel 50 meters. 5 m/s * 10 seconds = 50 meters. This is related to the concept of a rate because the velocity given is a rate. ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ Confidence Assessment: 3
.............................................
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): OK Self-critique rating: 3 ********************************************* 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 to we write the expression for the change `ds in the position? ********************************************* Your solution: Since vAve = ds/dt then ds = vAve*dt. ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ Confidence Assessment: 3
.............................................
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): OK Self-critique rating: 3 ********************************************* 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: Velocity, time interval, and displacement are related because velocity is a rate of change dependent on the time interval and displacement. ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ Confidence Assessment: 3
.............................................
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): OK Self-critique rating: 3 ********************************************* 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 this equation for ds, you would multiply each side by dt. This would give you ds = vAve * dt. ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ Confidence Assessment: 3
.............................................
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): OK Self-critique rating: 3 ********************************************* 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: The result shows that all three variables are inter-related. ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ Confidence Assessment: 2
.............................................
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): I wasn't sure as to what the question was asking until I read the given solution. Self-critique rating: 2 ********************************************* Question: `q013. What algebraic steps do we use to solve the equation vAve = `ds / `dt for `dt, and what is our result? ********************************************* Your solution: To solve for dt, you multiply both sides by dt, and then divide ds by vAve. This gives you dt = ds / vAve. ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ Confidence Assessment: 3
.............................................
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): OK Self-critique rating: 3 ********************************************* Question: `q014. How is this result related to our intuition about the meanings of the terms average velocity, displacement and clock time? ********************************************* Your solution: To find the time it takes to go a certain distance, you can divide the change in distance by the average velocity. ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ Confidence Assessment: 3
.............................................
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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK Self-critique rating: 3