#$&* course Phy 231 2/2 9:30 004. Acceleration
.............................................
Given Solution: The rate of change of the speed with respect clock time is equal to the change in the speed divided by the change in the clock time. So we must ask, what is the change in the speed, what is the change in the clock time and what therefore is the rate at which the speed is changing with respect to clock time? The change in speed from 5 meters/second to 25 meters/second is 20 meters/second. This occurs in a time interval lasting 4 seconds. The average rate of change of the speed is therefore (20 meters/second)/(4 seconds) = 5 meters / second / second. This means that on the average, per second, the speed changes by 5 meters/second. STUDENT QUESTION Would we not have s^2 since we are multiplying s by s? INSTRUCTOR RESPONSE That is correct. However in this question I've chosen not to confuse the issue by simplifying the complex fraction m/s/s, which we address separately. To clarify, m / s / s means, by the order of operations, (m/s) / s, which is (m/s) * (1/s) = m/s^2. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `q002. Explain in commonsense terms of the significance for an automobile of the rate at which its velocity changes. Do you think that a car with a more powerful engine would be capable of a greater rate of velocity change? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: If a car is able to change velocity quickly, that means it is able to speed up in a shorter amount of time compared to a car that is changing velocity more slowly, even if they are both speeding up. This would indicate a more powerful engine, because increasing speed at a faster rate means having enough power to more efficiently overcome other forces on the car (like wind resistance, friction on the road, etc.). (Alternatively, 'changing velocity quickly' could also mean slowing down very efficiently when necessary, which would mean really good brakes. Also important!) confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: A car whose velocity changes more rapidly will attain a given speed in a shorter time, and will be able to 'pull away from' a car which is capable of only a lesser rate of change in velocity. A more powerful engine, all other factors (e.g., weight and gearing) being equal, would be capable of a greater change in velocity in a given time interval. STUDENT COMMENT: The significance for an automobile of the rate at which its velocity changes is the amount of speed it takes to travel to a place in a certain amount of time. If one car is traveling along side another, and they are going to the same location the velocity will be how long it take this car to get to this location going at a speed other than the other car. If a car with a more powerful engine were to travel the same distance its velocity would be capable of a greater rate if increased speed occurred. INSTRUCTOR RESPONSE: It's necessary here to distinguish between velocity, which is a pretty intuitive concept, and rate of change of velocity, which is much less intuitive and less familiar. An object can change velocity at a constant rate, from rest to a very high velocity. All the while the rate of change of velocity with respect to clock time can be unchanging. So the rate of change of velocity with respect to clock time has nothing to do with how fast the object is moving, but rather with how quickly the velocity is changing. Moving from one location to another, the displacement is the change in position. If the displacement is divided by the time required we get the average rate of change of position with respect to clock time, or average velocity. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `q003. Explain how we obtain the units meters / second / second in our calculation of the rate of change of the car's speed. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Calculating rate of change of speed means measuring change in speed with respect to time. The units for this calcuation are as follows: Speed: measured in meters/second Time: measured in seconds To calculate change in speed with respect to time, we divide (change in speed) by (time). So, we divde (meters per second) by (seconds). Meters per second is itself an expression of division: meters over seconds, technically, since it looks at distance traveled over a specific amount of time. This leaves us with meters divided by seconds divided by seconds, or meters/sec/sec. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: When we divide the change in velocity, expressed in meters/second, by the duration of the time interval in seconds, we get units of (meters / second) / second, often written meters / second / second. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `q004. The unit (meters / second) / second is actually a complex fraction, having a numerator which is itself a fraction. Such a fraction can be simplified by multiplying the numerator by the reciprocal of the denominator. We thus get (meters / second) * (1/ second). What do we get when we multiply these two fractions? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: (Meters/second) * (1/second) = (meters * 1)/(seconds * seconds) = meters/seconds^2 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: Multiplying the numerators we get meters * 1; multiplying the denominators we get second * second, which is second^2. Our result is therefore meters * 1 / second^2, or just meters / second^2. If appropriate you may at this point comment on your understanding of the units of the rate of change of velocity. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `q005. If the velocity of an object changes from 10 m/s to -5 m/s during a time interval of 5 seconds, then at what average rate is the velocity changing with respect to clock time? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The total change in velocity is -15m/s (because it decreases in speed and eventually changes direction, going 5m/s the other way). The time interval in question is 5 seconds. So, the change in velocity is (-15m/s) per (5s) or (-15m/s)/(5s). This reduces to (-3m/s)/s, or -3m/s^2. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: We see that the velocity changes from 10 meters/second to -5 meters/second, a change of -15 meters / second, during a five-second time interval. A change of -15 m/s during a 5 s time interval implies an average rate of -15 m/s / (5 s) = -3 (m/s)/ s = -3 m/s^2. This is the same as (-3 m/s) / s, as we saw above. So the velocity is changing by -3 m/s every second. STUDENT QUESTION Do you have to do the step -3 m/s /s. Because I get the same answer not doing that. INSTRUCTOR RESPONSE Your solution read ' -5 m/s - 10 m/s = -15 m/s / 5 seconds = -3 m/s '. Everything was right except the units on your answer. So the answer to you question is 'Yes. It is very important to do that step.' The final answer in the given solution is '-3 m/s every second', which is not at all the same as saying just '-3 m/s'. -15 m/s / (5 s) = -3 m/s^2, which means -3 m/s per s or -3 cm/s every second or -3 m/s/s. -3m/s is a velocity. The question didn't ask for a velocity, but for an average rate of change of velocity. -3 m/s per second, or -3 m/s every second, or -3 m/s/s, or -3 m/s^2 (all the same) is a rate of change of velocity with respect to clock time. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): This is something that I am solid on mathematically, but have to strain to think about conceptually. We have a negative average acceleration here, but by the end of the time interval in question, the object is actually speeding up. It's simple enough here, but I'm certain this is something that is going to trip me up in more complicated problems. I think part of the problem is that it is hard to envision circumstances that would cause an object moving one direction to slow down at a steady rate until rest, then reverse direction and begin to speed up at exactly the same rate. But this problem doesn't require the rate of change to be steady, correct? Finding the average acceleration can be useful even if we don't have any reason to believe the rate is constant? ------------------------------------------------ Self-critique rating: 2
.............................................
Given Solution: The average rate would be expressed by [ave rate of velocity change with respect to clock time] = `dv / `dt. The expression [ave rate of velocity change with respect to clock time] is pretty cumbersome so we give it a name. The name we give it is 'average acceleration', abbreviated by the letter aAve. Using a to represent acceleration, write down the definition of average acceleration. The definition of average acceleration is aAve = `dv / `dt. Please make any comments you feel appropriate about your understanding of the process so far. STUDENT COMMENT: It’s average velocity so it would be aAve. INSTRUCTOR RESPONSE: Good, but note: It’s average acceleration (not average velocity) so it would be aAve. In most of your course acceleration is constant, so initial accel = final accel = aAve. In this case we can just use 'a' for the acceleration. STUDENT QUESTION If I understand this correctly, the average rate in which velocity changes is acceleration???? Where did average acceleration fit into the problem, the problem asked for the average rate that velocity changed? INSTRUCTOR RESPONSE Acceleration is rate of change of velocity with respect to clock time. So the terms 'average acceleration' and 'average rate of change of velocity with respect to clock time' are identical. The term 'average rate of change of velocity' actually leaves off the 'with respect to clock time', but in the context of uniformly accelerated motion 'average rate of change of velocity' is understood to mean 'average rate of change of velocity with respect to clock time' . &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): aAve= average acceleration= average rate of change in velocity with respect to change in clock time. It is interesting to me that most of the work in this course will deal with constant acceleration. Is that because we will be assuming ideal conditions for various forces? It is so hard to picture something uniformly speeding up and continuing to speed up indefinitely without anything interfering. ------------------------------------------------ Self-critique rating: 2
.............................................
Given Solution: We see that the runner's velocity changes from 6 meters/second to 9 meters/second, a change of `dv = 9 m/s - 6 m/s = 3 m/s, during a time interval lasting from t = 1.5 sec to t = 3.5 sec. The duration of the interval is `dt = 3.5 s - 1.5 s = 2.0 s. The rate at which the velocity changes is therefore 3 m/s / (2.0 s) = 1.5 m/s^2. STUDENT QUESTION I'm not understanding why you have the power of 2 for. INSTRUCTOR RESPONSE When you divide m/s by s you do the algebra of the fractions and get m/s^2. You don't get m/s. The distinction is essential: m/s^2 is a unit of acceleration. m/s is a unit of velocity. Velocity and acceleration are two completely different aspects of motion. The algebra of dividing m/s by s was given in a previous question in this document. In a nutshell, (m/s) / s = (m/s) * (1/s) = m/s^2. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `q008. On a graph of velocity vs. clock time, we can represent the two events of this problem by two points on the graph. The first point will be (1.5 sec, 6 meters/second) and the second point will be (3.5 sec, 9 meters / sec). What is the run between these points and what does it represent? What is the rise between these points what does it represent? What does the slope between these points what does it represent? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: We organize our graph with time on the x-axis and velocity on the y-axis. So, change in time will be tracked horizontally, and give us the ""run,"" which is the difference between the x values of two points. In this case, (3.5s-1.5)=2 sec. The run is 2s. Change in velocity will be tracked vertically, and give us the ""rise,"" which is the difference between the y values of two points. Here, (9m/s-6m/s)=3m/s. The rise is 3m/s. Slope is calculated by rise/run, or (3m/s)/2s, or 1.5m/s^2. This is our acceleration. So the slope on this graph of velocity v. time represents the acceleration of the object. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: The rise from the first point to the second is from 6 meters/second to 9 meters/second, or 3 m/s. This represents the change `dv in velocity. The run is from 1.5 seconds to 3.5 seconds, or 2 seconds, and represents the change `dt in clock time. The slope, being the rise divided by the run, is 3 m/s / (2 sec) = 1.5 m/s^2. This slope represents `dv / `dt, which is the average acceleration during the time interval. You may if you wish comment on your understanding to this point. STUDENT QUESTION Are we going to use the terms acceleration and average acceleration interchangeably in this course? I just want to make sure I understand. INSTRUCTOR RESPONSE Good question. The term 'acceleration' refers to instantaneous acceleration, the acceleration at a given instant. The term 'average acceleration' refers to the average acceleration during an interval, calculated by subtracting initial from final velocity and dividing by the change in clock time. If acceleration is uniform, it's always the same. If acceleration is uniform, then, it is unchanging. In that case the instantaneous acceleration at any instant is equal to the average acceleration over any interval. So when acceleration is uniform, 'acceleration' and 'average acceleration' are the same and can be used interchangeably. Acceleration isn't always uniform, so before using the terms interchangeably you should be sure you are in a situation where acceleration is expected to be uniform. This can be visualized in terms of graphs: The instantaneous acceleration can be represented by the slope of the line tangent to the graph of v vs. t, at the point corresponding to the specified instant. The average acceleration can be represented by the average slope between two points on a graph of v vs. t. If acceleration is uniform then the slope of the v vs. t graph is constant--i.e., the v vs. t graph is a straight line, and between any two points of the straight line the slope is the same. In this case the tangent line at a point on the graph is just the straight-line graph itself. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `q009. In what sense does the slope of any graph of velocity vs. clock time represent the acceleration of the object? For example, why does a greater slope imply greater acceleration? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: A graph of velocity v. time shows us how velocity is changing over a certain time interval. If an object is speeding up, the graph will be increasing, indicating a greater velocity each second than the previous second. If an object is moving at a constant rate, the graph will be a horizontal line, staying at the same velocity over the course of the whole time interval. If an object is slowing down, the graph will be decreasing, indicating a slower velocity each second than the previous second. If the slope, which represents acceleration, is very steep, it implies that velocity is changing a lot over the course of each second. In other words, on an steeply-increasing graph, an object is moving a whole lot faster each second than the previous second. A shallower slope implies less acceleration, a smaller change in speed per second. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: Since the rise between two points on a graph of velocity vs. clock time represents the change in `dv velocity, and since the run represents the change `dt clock time, the slope represents rise / run, or change in velocity /change in clock time, or `dv / `dt. This is the definition of average acceleration. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `q010. This is the same situation as in the preceding problem: An automobile coasts down a hill with a constant slope. At first its velocity increases at a very nearly constant rate. After it attains a certain velocity, air resistance becomes significant and the rate at which velocity changes decreases, though the velocity continues to increase. Describe a graph of velocity vs. clock time for this automobile (e.g., neither increasing nor decreasing; increasing at an increasing rate, a constant rate, a decreasing rate; decreasing at an increasing, constant or decreasing rate; the description could be different for different parts of the graph). YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: A velocity v. time graph would show, at first, a straight line with a positive slope, for the portion of the car's journey that involves a very nearly constant increase in velocity (acceleration). The line is straight because the velocity changes a consistent amount with each second for this interval. Then, at the point where air resistance becomes an issue, the straight line will become a curve that is still increasing, but at a decreasing rate. This portion will be a concave downward curve, although it will not actually bend downward during the duration in question, because velocity is still increasing. But, the slope will become less and less steep from that point on, indicating a negative change in acceleration. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: Your graph should have velocity as the vertical axis and clock time as the horizontal axis. The graph should be increasing since the velocity starts at zero and increases. At first the graph should be increasing at a constant rate, because the velocity is increasing at a constant rate. The graph should continue increasing by after a time it should begin increasing at a decreasing rate, since the rate at which the velocity changes begins decreasing due to air resistance. However the graph should never decrease, although as air resistance gets greater and greater the graph might come closer and closer to leveling off. Critique your solution by describing or insights you had or insights you had and by explaining how you now know how to avoid those errors. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I'm fairly confident, but not entirely certain, that it is appropriate to call this type of curve a ""concave downward"" one, since the curve never actually points downward. But my understanding is that this is the terminology we use in this situation. ------------------------------------------------ Self-critique rating:2
.............................................
Given Solution: Your graph should have acceleration as the vertical axis and clock time as the horizontal axis. At first the graph should be neither increasing nor decreasing, since it first the acceleration is constant. Then after a time the graph should begin decreasing, which indicates the decreasing rate at which velocity changes as air resistance begins having an effect. An accurate description of whether the graph decreases at a constant, increasing or decreasing rate is not required at this point, because the reasoning is somewhat complex and requires knowledge you are not expected to possess at this point. However it is noted that the graph will at first decrease more and more rapidly, and then less and less rapidly as it approaches the t axis. STUDENT QUESTION: Can you clarify some more the differences in acceleration and velocity? INSTRUCTOR RESPONSE: ** Velocity is the rate at which position changes and the standard units are cm/sec or m/sec. Acceleration is the rate at which velocity changes and its standard units are cm/s^2 or m/s^2. Velocity is the slope of a position vs. clock time graph. Acceleration is the slope of a velocity vs. clock time graph. ** STUDENT QUESTION: In the problem it states that velocity continues to increase even though the rate at which velocity changes decreases. I don’t understand your the slope will decrease if this is true. I can understand a diminish in velocity and time, but not a down turn of the slope, which is what your solution leans to. INSTRUCTOR RESPONSE Your thinking is good, but you need carefully identify what it is you're describing. The question here concerns the acceleration vs. clock time graph, whereas most of your comments apply to the velocity vs. clock time graph. Under these conditions the slope of the velocity vs. clock time graph will decrease, this will occur as long as the acceleration vs. clock time graph decreases, regardless of whether that decrease is at a constant, an increasing or a decreasing rate. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I see now that the curve may start out as concave downward, but it won't stay that way, as you describe it levelling off closer to the x-axis. Does that mean it will never technically reach 0, only approach it? ------------------------------------------------ Self-critique rating: