Some practice with velocity and position graphs

Transcription

1 Some practice with velocity and position graphs Position to Velocity The main idea here is that the velocity is the rate of change of the position. A large velocity means the position changes fast, a big change (up/down on the graph) in a little time(sideways on the graph), making a large slope on the graph. A small velocity means that there is very little change in position (up/down on the graph) in the same amount of time, making a small slope on the graph. So the velocity is shown by the slope of the line on the position graph (slope = rate of change). Thus to find velocity we look for the rise ( x) divided by the run ( t), giving us the velocity in m/s. For a graph with constant slopes this is fairly easy. The velocity is the same for all the points which have the same slope, making a set of flat lines : N S Each segment on the position graph gave a straight line on the velocity graph, with value = to slope. More complex curves can be thought of as made of line segments. Try this one: E6 W Of course a REAL position time graph does not have these sudden changes in velocity, it would be a smooth curve. Finding slopes for a curve is a bit more difficult, this is described in the next section!

2 Finding the slopes of a curve: Although an object may be changing its velocity, at any instant of time the object is moving with only one speed and in only one direction! Finding that speed and direction from a position vs. time graph means we have to find the slope of a curve at a point. Here the idea is that curves can be approximated (faked, sort of) as a series of short straight lines up close the curve is almost a line. Thus the closer one comes to an actual point the more a curve looks like a straight line. To see this imagine zooming in on a curve as shown below: The closer in you zoom, the more the curve looks like a straight line. The slope of the curve at a point is the slope of the straight line that the curve resembles/becomes. To find this line you could use calculus, which was designed by physicists to do this sort of thing, or you can use your own very powerful visual processor your brain! You are VERY good at seeing lines, so just free your processor to do that by hiding everything except a small segment of the line (I use my thumbs and a ruler like this: we want to find the slope here Use thumbs to hide the line except near the point. Position the ruler to match the line segment. Use ruler to draw a line extending the line segment so that its slope can be found The slope of the line you drew is the slope of the curve at that point! You know the line you drew is just right if you can t tell the difference between the line and the curve right at the point you are using! This line is called the tangent line. You don t need to find the tangent line at every point, just a few so you can connect them with a smooth line. Some things to watch for: Where is the slope zero (flat). When is it up or down (east or west, + or -, )? What is the steepest slope and what is the least steep? Where is the slope (velocity) roughly constant? Where is it changing quickly (very curved)? You can see an example of this next!

3 An Example Using a Curved Position-Time Graph These methods allow us to find approximate sketches quickly, or to make more detailed measurements and make a pretty good quality graph of velocity. v = changing v v make some tangents constant v v N S Some things to watch for: It is common to see velocity as more constant than it really is! A parabola on a position-time graph produces a straight line on a velocity-time graph. The amount of curve in the position-time graph tells you about the way the velocity-time graph will angle, more curve means more rapidly changing velocity which means a steeper slope on the velocity graph. Next we ll go backwards: from velocity to position. -

4 Velocity to Position If velocity is constant or we know the average velocity then getting the position is easy: we find the change in position from the velocity and time, and add up changes if necessary. s = v ave t Adding up the changes in position as time passes gives us the overall change in position up to that point. Eg: A person walks jogs east at. m/s for. seconds, then east at.m/s for.s, stops for.s and finally sprints east for.s at.m/s. On a velocity graph that looks like this: Motion East and West Times t v ave s Position( m) -s.s.m/s E. m E. m E -6s.s.m/s W m W 8. m W 6-9s.s.m/s m 8. m W 9-s.s.m/s E. m E. m W Motion East and West The positions we found are the positions at the END of each time interval, so we put them into our graph of position vs. time at these times. In this case the velocity is constant in between the steps, and we recall that a constant velocity on a position graph is a straight line segment. So that is how we link the points. When we get to curves it will be a little fancier.

5 As we saw in class, a graph where velocity changes suddenly between constant values is not very realistic. However if the graph is a straight line, or even a curve, we can turn it into such a graph by using the idea of average (effective) velocity. Consider something changing its velocity in a regular way: Motion with constant changes Motion with constant changes First we chop up the time into intervals. I have chosen -s, -s, -7s, 7-s. You could choose others, but too few and you don t get enough of a feel, too many and you do a lot of work you probably don t need to. I specifically choose to make one chop at 7.s because that is where the velocity is (a turn-around point). Now I find an average velocity for each region and a distance travelled: Times t v eff s total x Initial m (assumed) Position at s -s.s.m/s E.m E. m E -s.s.m/s E. m E 6. m E Position at s -7s.s.8m/s E.6 m E 7.8 m E Etc. 7-s.s. m/s W. m W. m E With curves we do the same thing, but making the horizontal line is a bit trickier. The secret is to get the same amount above and below the graph: Chop the time intervals: key points: max/min points, zero points. Place average velocity. This is too high, the upper part is smaller than the lower. This is too low, the lower part is smaller than the upper. This is just right, the two parts are equal. So v ave is read from the vertical axis.7m/s I call this the bow tie method, because getting the two halves equal is like making a bow tie work.

6 Some practice problems: Questions to refer to the following VELOCITY vs. TIME graphs: a) + b) + c) + d) + v(m/s ) - v(m/s ) -. Which of the graphs shows the largest displacement at the end of the time shown?. Which graph shows a constant velocity?. Which graph shows an object that changes direction during the time shown? Questions to 6 refer to the POSITION-TIME graphs below: (notice that these show something quite different than the first set (from questions to ) a) + b) + c) + d) + v(m/s ) - v(m/s ) - x(m) x(m) x(m) x(m) - -. Which graph or graphs show a constant velocity?. Which graph would have the greatest displacement at the end of the time shown? 6. Which graph shows an object that changes direction in the time shown? - - Questions to 8 refer to the following DISPLACEMENT vs. TIME graph for a moving car. x (km).e.e.e.w.w.w Position of Car t (hr) 7. At which of the following times does the car pass the (arbitrary) origin? a) hr b) hr c) 7hr d) 9hr e) never 8. When does the car have zero velocity? a) hr b) hr c) 7hr d) 9hr e) never 9. What is the approximate size of the average velocity of the car between hr and 7 hr? a) km/hr b) km/hr c) km/hr d) 7 km/hr e) km/hr. What is the size of the instantaneous velocity at s? a) km/hr b) km/hr c) 7 km/hr d) km/hr e) km/hr. What is the car s total displacement in the time shown? a) km b) km c) 6 km d) km e) 9 km. Between what times is the car going west? a) to 7 s b) to.7s c).7s to 9s d) after 7 s e) after 9 s Answers:

7 Solving Word Problems (extracting information) Examine the following word problem. Read it over, and try to imagine the motion. A car is stopped at a red light on Granville at No. Road. After. seconds the light turns green and the car moves east at m/s for the next. seconds. What is the car s average velocity over this entire time? ) In the problem above, circle the words that tell you about the cars motion. For example the fact that the car is stopped tells you about it's motion, so you should circle that word. There are others, some describing position, some velocity, some time. ) Now, next to each circle you made in question, write the variable symbol for the quality that it tells you about. For example the "stopped" tells you about a velocity, so write v v next to that circle. ) If there is more than one different velocity then give the velocities different names so you can tell them apart (like v v and v v ). Add the names to the variables you wrote beside the problems ) Now write down the velocity or velocities in a list, including direction. Use the space below. Make sure you include all the circled velocity information. INFO: vv = time v = time FIND: ) Write down the times that each of the velocities applies. Use the circled time information, and write the starting and ending time that the velocity applies 6) Draw a picture showing the car stopped at the light. Add another image of the car, showing where it moves (don't worry about numbers yet just a general idea). This is to help you be clearer about what is happening. 7) Add what you put into your list to the diagram. 8) Add what we are trying to find to the list above. Think about what that term means, and write down how one finds it (in this case this might be a formula. (If there is more than one way to find it then you might write down more than one thing, though at the moment I think there is just one definition for this problem). Make a note of what you need to know to find it. 9) You should see that you need to know the change in position over the whole time and the change in time. Now let's figure out how far the car has gone. Look at the first velocity given how long the car has this velocity, how far will it go in that time?. How long does the car have the second velocity in the problem? In that time how far will it go?. What is the total distance the car goes? What is the total time for the problem? ) You are pretty much done now! Write down the answer clearly.

8 Here is my version of doing this problem: A car is stopped at a red light on Granville at No. Road. After. seconds the light turns green and the car moves east at m/s for the next. seconds. What is the car s average velocity over this entire time? v =,.s INFO: v = m/s time =. s v = m/s E,.s v = m/s east time =. s s FIND: v ave SOL: CALC: Find s = v t, and the same for s. Find total s, t. v ave = s/ t s = m, s = ( m/s E)(.s) = 8 m E, so s total = 8 m E t total =.s +.s =. s v ave = 8 m E /.s = 9. m/s E Some word problems:. If you travel m N in. minutes, what is your effective velocity in m/s?. m/s N (. m/s N). If you run at 8. m/s for. minutes and then jog at. m/s for another minutes, how far have you gone?.7 km. What was your average (effective) velocity over the full time in the previous question?.8 m/s. You travelled at km/h for. minutes to get to Richmond centre, which is. km E of your reference point (your house). What was your initial position you must have begun travelling at? Either we started at.7 km W (going 6. km E) or we started at 7.7 km E (going 6. km E). You are running a race and travelling at 7.89 m/s from your coach's perspective. There is only person ahead of you! They are. m ahead to be exact. You catch up to the person in. s. What is the velocity of the person you caught up to, as measured by your coach, who is standing still watching the race from the sidelines?.9 m/s 6. Ms. Seminutin and I are going to run a race. Ms. Seminutin is much faster than me, since he can run at 8. m/s while I can only run at 6. m/s. If we are running a. m race how long a head start should Ms. Seminutin give me to make the race fair (to make us cross the finish line at close to the same time)? m or 8.7s 7. If Ms. Seminutin decided to give me a head start of 6. seconds in the previous question, at what time would he pass me? s after I start 8. How far from the finish line would we be? 6 m 9. A car is speeding down the highway at km/h when it passes a police car. The police car takes about. minutes to get started, but then travels at 6 km/h. How long does it take the police car to catch up to the speeder?. min. You are heading to the school, which is m north of where you are. You walk quite quickly, at about. m/s. Someone else is heading south away from the school at the same time. If you meet them when you are m from the school, what was their velocity?. m/s S