#$&* course phys 201 2/6 3:20 004. Acceleration
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Given Solution: The rate of change of the speed with respect clock time is equal to the change in the speed divided by the change in the clock time. So we must ask, what is the change in the speed, what is the change in the clock time and what therefore is the rate at which the speed is changing with respect to clock time? The change in speed from 5 meters/second to 25 meters/second is 20 meters/second. This occurs in a time interval lasting 4 seconds. The average rate of change of the speed is therefore (20 meters/second)/(4 seconds) = 5 meters / second / second. This means that on the average, per second, the speed changes by 5 meters/second. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Ok ------------------------------------------------ Self-critique rating: 3 ********************************************* Question: `q002. Explain in commonsense terms of the significance for an automobile of the rate at which its velocity changes. Do you think that a car with a more powerful engine would be capable of a greater rate of velocity change? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I do think a car with a more powerful engine would be capable of a greater rate of velocity of acceleration. If a car has a bigger engine that means it has more force to push the are to accelerate faster making it able to change the velocity at a faster rate. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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. Self-critique (if necessary): Ok Self-critique rating: 3 ********************************************* 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: In terms of the cars speed, we can obtain this information by (meters / sec) / (meters / sec/ sec )= ((m/sec) *(1/(m/sec))) / sec = m/sec/sec confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: When we divide the change in velocity, expressed in meters/second, by the duration of the time interval in seconds, we get units of (meters / second) / second, often written meters / second / second. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: 2 ********************************************* Question: `q004. The unit (meters / second) / second is actually a complex fraction, having a numerator which is itself a fraction. Such a fraction can be simplified by multiplying the numerator by the reciprocal of the denominator. We thus get (meters / second) * (1/ second). What do we get when we multiply these two fractions? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: When we multiply m by 1 it give us m. Then we multiply the denominator of s by s giving us s^2. Put the numerator and denominator together and we get (meters / sec2). confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): Ok ------------------------------------------------ Self-critique rating: 3 ********************************************* Question: `q004. 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: First, you subtract final minus initial object changes: (-5 - (10)), then divide by the time :5 sec = -3 m/s2 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): Ok ------------------------------------------------ Self-critique rating: 3 ********************************************* Question: `q005. You should have noted that velocity can be positive or negative, as can the change in velocity or the rate at which velocity changes. The average rate at which a quantity changes with respect to time over a given time interval is equal to the change in the quantity divided by the duration of the time interval. In this case we are calculating the average rate at which the velocity changes. If v represents velocity then we we use `dv to represent the change in velocity and `dt to represent the duration of the time interval. What expression do we therefore use to express the average rate at which the velocity changes? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: We would use the expression aAve= dv / dt confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Ok ------------------------------------------------ Self-critique rating: 3 ********************************************* Question: `q006. If a runner is moving at 6 meters / sec at clock time t = 1.5 sec after starting a race, and at 9 meters / sec at clock time t = 3.5 sec after starting, then what is the average acceleration of the runner between these two clock times? If you can, answer the question as posed. If not, first consider the two questions below: What is the change `dv in the velocity of the runner during the time interval, and what is the change `dt in clock time during this interval? What therefore is the average rate at which the velocity is changing with respect to clock time during this time interval? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: First, subtract the final running speed by the initial running speed to find the change in running speed. Then you would divide this answer by the change in clock times (final -initial times) Ave = (9m/s - 6 m/s) / (3.5s - 1.5s) = (3m/s) / (2m/s) = 1.5 m/s/s confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: We see that the runner's velocity changes from 6 meters/second to 9 meters/second, a change of `dv = 9 m/s - 6 m/s = 3 m/s, during a time interval lasting from t = 1.5 sec to t = 3.5 sec. The duration of the interval is `dt = 3.5 s - 1.5 s = 2.0 s. The rate at which the velocity changes is therefore 3 m/s / (2.0 s) = 1.5 m/s^2. Self-critique (if necessary): Ok ------------------------------------------------ Self-critique rating: 3 ********************************************* Question: `q007. On a graph of velocity vs. clock time, we can represent the two events of this problem by two points on the graph. The first point will be (1.5 sec, 6 meters/second) and the second point will be (3.5 sec, 9 meters / sec). What is the run between these points and what does it represent? What is the rise between these points what does it represent? What does the slope between these points what does it represent? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: the slope represents the acceleration between the two points= rise / run = (y2 - y1) / (x2 - x1) = 3 / 2 Run = 2 sec ; represents the change in clock time Rise = 3 sec ; represents the change velocity confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): Ok ------------------------------------------------ Self-critique rating: 3 ********************************************* Question: `q008. 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: Acceleration = aAve = dv / dt = (y2 - y1) / (x2 - x1) , this is the same equation for the slope in which case they are the same. The greater the slope signifies a greater acceleration because if they are the same formula they are going to change proportionality to each other. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Since the rise between two points on a graph of velocity vs. clock time represents the change in `dv velocity, and since the run represents the change `dt clock time, the slope represents rise / run, or change in velocity /change in clock time, or `dv / `dt. This is the definition of average acceleration. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Ok ------------------------------------------------ Self-critique rating: 3 ********************************************* Question: `q009. This is the same situation as in the preceding problem: An automobile coasts down a hill with a constant slope. At first its velocity increases at a very nearly constant rate. After it attains a certain velocity, air resistance becomes significant and the rate at which velocity changes decreases, though the velocity continues to increase. Describe a graph of velocity vs. clock time for this automobile (e.g., neither increasing nor decreasing; increasing at an increasing rate, a constant rate, a decreasing rate; decreasing at an increasing, constant or decreasing rate; the description could be different for different parts of the graph). YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Towards the beginning of the graph, there would be a steep line representing the fast acceleration at a constant rate. Then maybe toward the middle of the line, the line would be being to not be as steep but still be accelerating representing the acceleration but at a decreasing rate. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): Ok ------------------------------------------------ Self-critique rating: 3 ********************************************* Question: `q010. 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 acceleration 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: Towards the beginning of the graph the line would be increasing in a positive direction in a straight line. However, after wind becomes a significant factor, the line would still be accelerating but at a slower rate also the line would be less steep. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Your graph should have acceleration as the vertical axis and clock time as the horizontal axis. At first the graph should be neither increasing nor decreasing, since it first the acceleration is constant. Then after a time the graph should begin decreasing, which indicates the decreasing rate at which velocity changes as air resistance begins having an effect. An accurate description of whether the graph decreases at a constant, increasing or decreasing rate is not required at this point, because the reasoning is somewhat complex and requires knowledge you are not expected to possess at this point. However it is noted that the graph will at first decrease more and more rapidly, and then less and less rapidly as it approaches the t axis. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Ok ------------------------------------------------ Self-critique rating: 3 Question: `q012. What do we get if we divide 40 meters by 5 meters / second? Your answer will include a number and its units. You should explain how you got the units of your answer. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 40 m / 5 m/sec = the meters cancel out leaving us with 8 sec. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ********************************************* Question: `q013. On a graph of velocity v vs. clock time t we have the two points (5 s, 10 m/s) and (10 s, 20 m/s). What is the average slope of the graph between these points, and what does this average slope mean? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Average slope means the average between the two points of velocity. (20 m/s - 10 m/s) / (10 sec - 5 sec) = 10 m/s / 5 sec " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: Question: `q012. What do we get if we divide 40 meters by 5 meters / second? Your answer will include a number and its units. You should explain how you got the units of your answer. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 40 m / 5 m/sec = the meters cancel out leaving us with 8 sec. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ********************************************* Question: `q013. On a graph of velocity v vs. clock time t we have the two points (5 s, 10 m/s) and (10 s, 20 m/s). What is the average slope of the graph between these points, and what does this average slope mean? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Average slope means the average between the two points of velocity. (20 m/s - 10 m/s) / (10 sec - 5 sec) = 10 m/s / 5 sec " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!