#$&* course Phy 231 2/5 10:15pm 006. Using equations with uniformly accelerated motion.
.............................................
Given Solution: The equation vf = v0 + a * `dt is solved for a by first adding -v0 to both sides to obtain vf - v0 = v0 + a * `dt - v0, which simplifies to vf - v0 = a * `dt. Both sides are then divided by `dt to obtain (vf - v0) / `dt = a. Reversing left-and right-hand sides we obtain the formula a = (vf - v0) / `dt. We then plug in our given values of initial and final velocities and the time interval. Since velocity increases from 10 m/s to 30 m/s, initial velocity is v0 = 10 m/s and final velocity is vf = 30 m/s. The time interval `dt is 15 seconds, so we have a = (30 m/s - 10 m/s) / (15 s) = 20 m/s / (15 s) = 1.33.. m/s^2. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `q002. It wasn't necessary to use a equation to solve this problem. How could this problem had been reasoned out without the use of an equation? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Looking at the initial and final velocities, we can plainly see that the velocity changed by 20m/s. The problem states this took 15s, so we already have everything we need to understand what is happening in terms of a speed changing by a certain amount in a certain time. That amount is 20m/s and that time is 15s. We know we need to divide, because that 20m/s change happened in a series of 15 one-second intervals, so we need to divide 20 by 15. More specifically, 20m/s divided by 15s. Some quick simplification tells us this reduces to 4m/s/3s, or 1.333 m/s^2. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: Knowing that acceleration is the average rate at which velocity changes, we would first find the change in velocity from 10 meters/second to 30 meters/second, which is 20 meters/second. We would then divided change in velocity by the time interval to get 20 meters/second / (15 sec) = 1.33 m/s^2. STUDENT QUESTION (about reasoning vs. using the equation) I understand but the steps taken to get to the acceleration were the steps of the equation????? INSTRUCTOR RESPONSE The steps outlined here are the steps we could use to derive the equation. However it's possible to use the equation blindly, without understanding the reasoning behind it. In fact this is how most student use the equation, if not asked questions of this nature about the reasoning. So, this question asks for the reasoning. The first statement in the given solution is 'Knowing that acceleration is the average rate at which velocity changes, we would first find the change in velocity from 10 meters/second to 30 meters/second, which is 20 meters/second.' When using the equation you never explicitly find or reason out the change in velocity, though of course the change in velocity is there in the equation, represented by the term a * `dt. In other words, you do find it, but you can use the equation without ever recognizing that you have done so. Similarly the step a = (30 m/s - 10 m/s) / 15 s in your equation-based solution does correctly divide the change in velocity by the time interval, but you can use the equation to do this without ever recognizing that you have done so. The direct reasoning solution never mentions or uses the equation, though of course direct reasoning can be used to derive the equation. This should help illustrate the difference between direct reasoning and using an equation. Both skills are important. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `q003. Use the equation `ds = (vf + v0) / 2 * `dt to determine the initial velocity of an object which accelerates uniformly through a distance of 80 meters in 10 seconds, ending up at a velocity of 6 meters / sec. begin by solving the equation for the desired quantity. Show every step of your solution. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: `ds = (vf+v0)/2 * `dt divide both sides by `dt: (`ds)/(`dt) = (vf + v0)/2 multiply both sides by 2: 2(`ds)/(`dt)=(vf+v0) subtract vf from both sides 2(`ds)/(`dt)-vf=v0 2(80m)/10s - 6m/s = v0 160/10s - 6m/s = v0 16m/s - 6m/s = v0 10m/s = v0 Initial velocity is 10m/s confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: We begin by solving the equation for v0. Starting with `ds = (vf + v0) / 2 * `dt, we can first multiply both sides of the equation by 2 / `dt, which gives us `ds * 2 / `dt = (vf + v0) / 2 * `dt * 2 / `dt. The right-hand side can be rearranged to give (vf + v0) * `dt / `dt * 2 / 2; since `dt / `dt = 1 and 2 / 2 = 1 the right-hand side becomes just vf + v0. The equation therefore becomes 2 * `ds / `dt = vf + v0. Adding -vf to both sides we obtain v0 = 2 * `ds / `dt - vf. We now plug in `ds = 80 meters, `dt = 10 sec and vf = 6 m/s to get v0 = 2 * 80 meters / 10 sec - 6 m/s = 160 meters / 10 sec - 6 m/s = 16 m/s - 6 m/s = 10 m/s. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `q004. We can reconcile the above solution with straightforward reasoning. How could the initial velocity have been reasoned out from the given information without the use of an equation? Hint: two of the quantities given in the problem can be combined to give another important quantity, which can then be combined with the third given quantity to reason out the final velocity. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Knowing that the object travels 80 meters in 10 seconds, we can quickly determine that the average velocity during that interval is 8m/s. Now, we compare average velocity to the given final velocity of 6m/s, and realize it must have started out at 10m/s in order to attain an average of 8m/s. So, the initial velocity must have been 10m/s. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: The average velocity of the object is the average rate at which its position changes, which is equal to the 80 meters change in position divided by the 10 s change in clock time, or 80 meters / 10 sec = 8 meters / sec. Since the 8 m/s average velocity is equal to the average of the unknown initial velocity and the 6 m/s final velocity, we ask what quantity when average with 6 m/s will give us 8 m/s. Knowing that the average must be halfway between the two numbers being averaged, we see that the initial velocity must be 10 m/s. That is, 8 m/s is halfway between 6 m/s and 10 m/s. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `q005. Using the equation `ds = v0 `dt + .5 a `dt^2 determine the initial velocity of an object which accelerates uniformly at -2 m/s^2, starting at some unknown velocity, and is displaced 80 meters in 10 seconds. Begin by solving the equation for the unknown quantity and show every step. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: `ds = v0`dt + .5a `dt^2 Subtract .5a`dt^2 from both sides: `ds-.5a`dt^2 = v0`dt Divide both sides by `dt: (`ds - .5a`dt^2)/`dt = v0 (80m - .5(-2m/s^2)(10s)^2)/10s = v0 (80m + (1m/s^2)(100s^2))/10s = v0 (80m + 100m*s^2/s^2)/10s = v0 (80m + 100m)/10s = v0 180m/10s = v0 18m/s = v0 Initial velocity is 18m/s confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: The unknown quantity is the initial velocity v0. To solve for v0 we start with `ds = v0 `dt + .5 a `dt^2. We first add -.5 a `dt^2 to both sides to obtain `ds - .5 a `dt^2 = v0 `dt. We then divide both sides by `dt to obtain (`ds - .5 a `dt^2) / `dt = v0. Then we substitute the given displacement `ds = 80 meters, acceleration a = -2 m/s^2 and time interval `dt = 10 seconds to obtain v0 = [ 80 meters - .5 * (-2 m/s^2) * (10 sec)^2 ] / (10 sec) = [ 80 meters - .5 * (-2 m/s^2) * 100 s^2 ] / (10 sec) = [ 80 m - (-100 m) ] / (10 sec) = 180 m / (10 s) = 18 m/s. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `q006. Check the consistency of this result by verifying, by direct reasoning rather than equations, that an object whose initial velocity is 18 m/s and which accelerates for 10 seconds at an acceleration of -2 m/s^2 does indeed experience a displacement of 80 meters. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Accelerating by -2m/s^2 for 10 seconds means a total change in velocity of -20m/s. In this case, that means the object slows from 18m/s all the way to rest and then begins traveling in the opposite direction, reaching a spead of -2m/s (that is, 2m/s in the other direction) by the end of the 10-second interval. This doesn't stop us from finding the average velocity which is halfway between initial velocity of 18m/s and final velocity of -2m/s. That average velocity is 8m/s. With that 8m/s in mind, we look again at the time interval of 10 seconds and can reason that the object traveled a total of 80m during the interval. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: The change in the velocity of the object will be -2 m/s^2 * 10 s = -20 m/s. The object will therefore have a final velocity of 18 m/s - 20 m/s = -2 m/s. Its average velocity will be the average (18 m/s + (-2 m/s) ) / 2 = 8 m/s. An object which travels at an average velocity of 8 m/s for 10 sec will travel 80 meters. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `q007. Using the equation vf^2 = v0^2 + 2 a `ds determine the initial velocity of an object which attains a final velocity of 20 meters/second after accelerating uniformly at 2 meters/second^2 through a displacement of 80 meters. Begin by solving the equation for the unknown quantity and show every step. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: vf^2 = v0^2 + 2a`ds subtract 2a`ds from both sides vf^2-2a`ds = v0^2 take the square root of both sides sqrt (vf^2-2a`ds) = v0 sqrt [(20m/s)^2 - 2(2m/s^2)(80m)] = v0 sqrt (400m^2/s^2 - 320m^2/s^2) = v0 sqrt (80m^2/s^2) = v0 sqrt80 m/s = v0 initial velocity is approx. 8.94m/s confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: To solve for the unknown initial velocity v0 we start with vf^2 = v0^2 + 2 a `ds. We first add -2 a `ds to both sides to obtain vf^2 - 2 a `ds = v0^2. We then reverse the right-and left-hand sides and take the square root of both sides, obtaining v0 = +- `sqrt( vf^2 - 2 a `ds). We then substitute the given quantities vf = 20 m/s, `ds = 80 m and a = 3 m/s^2 to obtain v0 = +- `sqrt( (20 m/s)^2 - 2 * 2 m/s^2 * 80 m) = +- `sqrt( 400 m^2 / s^2 - 320 m^2 / s^2) = +- `sqrt(80 m^2 / s^2) = +- 8.9 m/s (approx.). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Oops, forgot the +-. That will take some focus to train myself to look out for. ------------------------------------------------ Self-critique rating: ********************************************* Question: `q008. We can verify that starting at +8.9 m/s an object which attains a final velocity of 20 m/s while displacing 80 meters must accelerate at 2 m/s^2. In this case the average velocity will be ( 8.9 m/s + 20 m/s) / 2 = 14.5 m/s (approx) and the change in velocity will be 20 m/s - 8.9 m/s = 11.1 m/s. At average velocity 14.5 meters/second the time required to displace the 80 meters will be 80 m / (14.5 sec) = 5.5 sec (approx). The velocity change of 11.1 meters/second in 5.5 sec implies an average acceleration of 11.1 m/s / (5.5 sec) = 2 m/s^2 (approx), consistent with our results. Verify that starting at -8.9 m/s the object will also have acceleration 2 meters/second^2 if it ends up at velocity 20 m/s while displacing 80 meters. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: If given -8.9m/s as the starting velocity and and 20m/s as the final velocity, we can see that the average velocity, or halfway point between those velocities, is about 5.55m/s. We know it traveled 80 meters at that average velocity, so to find how many seconds passed we need to divide 80 meters by the amount it traveled per second: so, 80m/5.55m/s, for a result of about 14.41 seconds. Now we need to check how fast the speed of the object would have to change in order to get there in 14.41 seconds. Looking again at intial and final velocities, we can see that the total change in velocity was 28.9m/s. Since that change took place over the course of 14.41 seconds, we can divide 28.9m/s by 14.41s to figure out how much, per second, speed would have to increase. (28.9m/s)/(14.414s)=approximately 2 m/s^2. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: In this case the average velocity will be ( -8.9 m/s + 20 m/s) / 2 = 5.5 m/s (approx) and the change in velocity will be 20 m/s - (-8.9 m/s) = 28.9 m/s (approx). At average velocity 5.5 meters/second the time required to displace the 80 meters will be 80 m / (5.5 sec) = 14.5 sec (approx). The velocity change of 28.5 meters/second in 14.5 sec implies an average acceleration of 28.5 m/s / (14.5 sec) = 2 m/s^2 (approx), again consistent with our results. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `q009. Describe in commonsense terms the motion of the object in this example if its initial velocity is indeed -8.9 m/s. Assume that the object starts at the crossroads between two roads running North and South, and East and West, respectively, and that the object ends up 80 meters North of the crossroads. In what direction does it start out, what happens to its speed, and how does it end up where it does? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Since we know that the eventual positive displacement puts the object North of its starting point, then we can reason that beginning with a negative velocity means traveling in the opposite direction of its final position. So, it is starting out by traveling south of the crossroads. At the instant it begins traveling south, its velocity is -8.9m/s. That is, if the velocity wasn't constantly changing, it would travel 8.9m south in the first second. But, the velocity is becoming more and more positive by 2m/s per second. In this case, that means it is approaching zero, or getting slower. So by the end of the first second, it is only traveling 6.9m/s southbound. Eventually, its southbound velocity will reach 0. It will come to rest somewhere south of the crossroads, briefly, but then its speed will continue to get more and more positive, so it will actually reverse direction at that point and start traveling northbound, continuing to speed up by 2m/s per second as it goes. We know that total displacement is 80 meters north of the crossroads, but in fact the object will travel a greater distance than that if you add up its initial southbound travel, pivot north, return to the crossroads, and THEN 80-meter journey north to its eventual destination. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: The object ends up at position +80 meters, which is assumed to be 80 meters to the North. Its initial velocity is -8.9 m/s, with the - sign indicating that the initial velocity is in the direction opposite to the displacement of the object. So the object must start out moving to the South at 8.9 meters/second. Its acceleration is +2 m/s^2, which is in the opposite direction to its initial velocity. This means that the velocity of the object changes by +2 m/s every second. After 1 second the velocity of the object will therefore be -8.9 m/s + 2 m/s = -6.9 m/s. After another second the velocity will be -6.9 m/s + 2 m/s = -4.9 m/s. After another second the velocity will be -2.9 m/s, after another -.9 m/s, and after another -.9 m/s + 2 m/s = +1.1 m/s. The speed of the object must therefore decrease, starting at 8.9 m/s (remember speed is always positive because speed doesn't have direction) and decreasing to 6.9 m/s, then 4.9 m/s, etc. until it reaches 0 for an instant, and then starts increasing again. Since velocities after that instant become positive, the object will therefore start moving to the North immediately after coming to a stop, picking up speed at 2 m/s every second. This will continue until the object has attained a velocity of +20 meters/second and has displaced +80 meters from its initial position.{}{}It is important to understand that it is possible for velocity to be in one direction and acceleration in the other. In this case the initial velocity is negative while the acceleration is positive. If this continues long enough the velocity will reach zero, then will become positive. STUDENT QUESTION I understood the negative velocity but was unsure how to explain the rest. I am still rather confused by the last paragraph, expecially where it says that it is possible for velocity to be in one direction and acceleration in the other. INSTRUCTOR RESPONSE If you speed up the acceleration is in the direction of motion. If you slow down the acceleration is opposite the direction of motion. To speed up a wagon you can get behind it and push in the direction of its motion, giving it an acceleration in its direction of motion. To slow it down you can get in front of it and push it against its direction of motion (not advisable if it's a big wagon; think of stopping a child in a small wagon), giving it an acceleration in the direction opposite its motion. STUDENT COMMENTS Made relative sense, but still unsure in doubt to my answer that the object was increasing after it was moving in the right direction. Being negative, would it have started towards the south at that acceleration, and moving north would it have diminished its negativity? It seems this naturally. INSTRUCTOR RESPONSE Accelerating to the north (not 'moving' to the north; it does end up moving to the north, but the object starts out moving to the south), the speed in the southward direction would have diminished, as you say. Eventually it comes to rest, just for an instant, somewhere south of its starting point. Then the northward acceleration will give it an increasing northward velocity. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: If you understand the assignment and were able to solve the previously given problems from your worksheets, you should be able to complete most of the following problems quickly and easily. If you experience difficulty with some of these problems, you will be given notes and we will work to resolve difficulties. ********************************************* Question: `q010. An object speeds up from 10 m/s to 20 m/s, accelerating uniformly and traveling 60 meters during this interval. Specify which of the quantities v_0, v_f, aAve, `ds and `dt are given, and specify the value of each. Specify which of the four equations of uniformly accelerated motion include the given three quantities. There is at least one such equation, and there might be two. For each of the equations you specified, identify the quantity for which the value is not given. Then symbolically solve the equation for each of these quantities, showing the steps of your algebra. Substitute the three given quantities into your solution, and simplify. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: v0: 10m/s vf: 20m/s `ds: 60m We can use these quantities in the following formulas `ds = (v0 + vf) / 2 * `dt (to solve for `dt) vf^2 = v0^2 + 2 a `ds (to solve for a) `ds = (v0+vf)/2 * `dt multiply each side by 2: 2`ds = (v0+vf)*`dt divide each side by (v0+vf) (2`ds)/(v0+vf)=`dt (2 * 60m)/(10m/s+20m/s)=`dt (120m)/(30m/s)=`dt 40s=`dt Time interval is 40 seconds. vf^2 = v0^2 + 2a`ds subtract v0^2 from each side vf^2-v0^2 = 2a`ds divide each side by 2`ds (vf^2-v0^2)/(2`ds) = a [(20m/s)^2-(10m/s)^2]/(2*60m) = a (400m^2/s^2-100m^2/s^2)/(120m) = a (300m^2/s^2)/120m = a 2.5m/s^2 = a Acceleration is 2.5m/s^2. ********************************************* Question: `q011. An cart initially moving at 10 cm/s travels 40 cm while accelerating at 5 cm/s^2. Using the equations of uniformly accelerated motion determine the time required and the the cart's final velocity. Hint: You will need to start out with either the third or the fourth equation of uniformly accelerated motion. You are advised that it's easier to start out with the fourth equation. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: v0: 10cm/s `ds: 40cm a: 5cm/s^2 vf^2 = v0^2 + 2 a `ds. vf = sqrt (v0^2 + 2a`ds) vf = sqrt [(10cm/s)^2 + 2(5cm/s^2)(40cm)] vf = sqrt (100cm^2/s^2 + 400cm^2/s^2) vf = sqrt (500cm^2/s^2) vf = 22.4cm/s Final velocity is 22.4cm/s `ds = (v0 + vf) / 2 * `dt Multiply each side by 2 2`ds = (v0+vf) * `dt Divide each side by (v0+vf) (2`ds)/(v0+vf) = `dt (2*40cm)/(10cm/s + 22.4cm/s) = `dt (80cm)/(32.4cm/s)= 2.47s The time required is 2.47s. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ********************************************* Question: `q012. An object starts at position x = +10 cm with a velocity of +5 cm/s, and accelerates uniformly, ending up at position x = -30 cm after a time interval of 8 seconds. What is its velocity at this point, and what was its acceleration during this interval? Principles of Physics students should not spend over 5 minutes on this problem, General College Physics students should not spend over 10 minutes. University Physics students are expected to be able to solve this problem, but if it hasn't been solved within 15 minutes, should submit their best thinking and await the instructor's notes. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: v0: 5cm/s `ds: -30cm - 10cm = -40cm `dt: 8s `ds = (v0 + vf) / 2 * `dt 2(`ds) = (v0+vf)*`dt 2(`ds)/(`dt) = v0+vf vf = 2(`ds)/(`dt) - v0 vf= 2(-40cm)/(8s) - 5cm/s vf= -10cm/s - 5cm/s vf= -15cm/s Final velocity is -15cm/s. a=(vf-v0)/`dt a=(-15cm/s-5cm/s)/8s a=(-20cm/s)/8 a=-2.5cm/s^2 Acceleration is -2.5cm/s^2. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `q013. (University Physics only) The velocity of an object is given by the function v(t) = -.2 m/s^2 * t + 10 m/s. For the interval between t = 10 seconds and t = 30 seconds, find v_0 = v(10 sec) and v_f = v(30 sec). Then use direct reasoning with your known to determine how far the object travels during this interval. Integrate the velocity function to find a change-in-position function x(t) and show that this function predicts the same change in position as determined by direct reasoning. What is the position function x(t) if it is known that x(0 sec) = 50 meters? For this function, what is x(30) - x(10) and what is the significance of this result? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: v_0 = v(10sec) = -.2m/s^2 * 10s + 10m/s =-2m/s + 10m/s =8m/s Initial velocity is 8m/s v_f = v(30sec) = -.2m/s^2 * 30s + 10m/s =-6m/s + 10m/s =4m/s Final velocity is 4m/s From our calculated initial and final velocities we can determine that average velocity is 6m/s. For the given duration of 20s, that works out to 120m traveled during this interval. Taking away units, which is the only way I can figure out how to integrate this! (???Is it possible to keep units around while we integrate???)