端木·宇 2008-6-19 20:23
Kinematics with Graphs
Since you are not allowed to use calculators, SAT II Physics places aheavy emphasis on qualitative problems. A common way of testingkinematics qualitatively is to present you with a graph plottingposition vs. time, velocity vs. time, or acceleration vs. time and toask you questions about the motion of the object represented by thegraph. Because SAT II Physics is entirely made up of multiple-choicequestions, you won’t need to know how to draw graphs; you’ll just haveto interpret the data presented in them.
Knowing how to read such graphs quickly andaccurately will not only help you solve problems of this sort, it willalso help you visualize the often-abstract realm of kinematicequations. In the examples that follow, we will examine the movement ofan ant running back and forth along a line.
[align=center][img]http://www.24en.com/d/file/sat/sat2/physics/2008-01-24/7525d83a5b0475bea3b202d6d81c6314.gif[/img][/align]
[b] Position vs. Time Graphs[/b]
Position vs. time graphs give you an easy andobvious way of determining an object’s displacement at any given time,and a subtler way of determining that object’s velocity at any giventime. Let’s put these concepts into practice by looking at thefollowing graph charting the movements of our friendly ant.
[align=center][img]http://www.24en.com/d/file/sat/sat2/physics/2008-01-24/95989f0c457414b30ae522bf9d2554a8.gif[/img][/align]
Any point on this graph gives us the position of the ant at a particular moment in time. For instance, the point at (2,–2)tells us that, two seconds after it started moving, the ant was twocentimeters to the left of its starting position, and the point at (3,1) tells us that, three seconds after it started moving, the ant is one centimeter to the right of its starting position.
Let’s read what the graph can tell us aboutthe ant’s movements. For the first two seconds, the ant is moving tothe left. Then, in the next second, it reverses its direction and movesquickly to [i]y[/i] = 1. The ant then stays still at [i]y[/i] = 1for three seconds before it turns left again and moves back to where itstarted. Note how concisely the graph displays all this information.
[b] Calculating Velocity[/b]
We know the ant’s displacement, and we knowhow long it takes to move from place to place. Armed with thisinformation, we should also be able to determine the ant’s velocity,since velocity measures the rate of change of displacement over time.If displacement is given here by the vector [i][b]y[/b][/i], then the velocity of the ant is
[align=center][img]http://www.24en.com/d/file/sat/sat2/physics/2008-01-24/1348dcc4b06e9c5aa0d211694476c47c.gif[/img][/align] If you recall, the slope of a graph is a measure of rise over run; that is, the amount of change in the [i]y[/i] direction divided by the amount of change in the [i]x[/i] direction. In our graph, [img]http://www.24en.com/d/file/sat/sat2/physics/2008-01-24/2959e7b45f2ec13e7f11acbcdd7bf2cf.gif[/img] is the change in the [i]y[/i] direction and [img]http://www.24en.com/d/file/sat/sat2/physics/2008-01-24/72b94a8f3b486add5e4a1adf368f6d04.gif[/img] is the change in the [i]x[/i] direction, so [i][b]v[/b][/i] is a measure of the slope of the graph. [i]For any position vs. time graph, the velocity at time [/i]t[i] is equal to the slope of the line at [/i]t[i].[/i]In a graph made up of straight lines, like the one above, we can easilycalculate the slope at each point on the graph, and hence know theinstantaneous velocity at any given time.
We can tell that the ant has a velocity of zero from [i]t[/i] = 3 to [i]t [/i]= 6,because the slope of the line at these points is zero. We can also tellthat the ant is cruising along at the fastest speed between [i]t [/i]= 2 and [i]t[/i] = 3,because the position vs. time graph is steepest between these points.Calculating the ant’s average velocity during this time interval is asimple matter of dividing rise by run, as we’ve learned in math class.
[align=center][img]http://www.24en.com/d/file/sat/sat2/physics/2008-01-24/0942d6e55708787e915fe8ab15042e06.gif[/img][/align]
[b] Average Velocity[/b]
How about the average velocity between [i]t [/i]= 0 and [i]t [/i]= 3? It’s actually easier to sort this out with a graph in front of us, because it’s easy to see the displacement at [i]t [/i]= 0 and [i]t [/i]= 3, and so that we don’t confuse displacement and distance.
[align=center][img]http://www.24en.com/d/file/sat/sat2/physics/2008-01-24/adbafc010694688fd69d2eaf9e2577e8.gif[/img][/align][b] Average Speed[/b]
Although the total displacement in the firstthree seconds is one centimeter to the right, the total distancetraveled is two centimeters to the left, and then three centimeters tothe right, for a grand total of five centimeters. Thus, the averagespeed is not the same as the average velocity of the ant. Once we’vecalculated the total distance traveled by the ant, though, calculatingits average speed is not difficult:
[align=center][img]http://www.24en.com/d/file/sat/sat2/physics/2008-01-24/96eabd4a007eb92e8948a622e8fbefe0.gif[/img][/align]
[b] Curved Position vs. Time Graphs[/b]
This is all well and good, but how do youcalculate the velocity of a curved position vs. time graph? Well, thebad news is that you’d need calculus. The good news is that SAT IIPhysics doesn’t expect you to use calculus, so if you are given acurved position vs. time graph, you will only be asked qualitativequestions and won’t be expected to make any calculations. A few pointson the graph will probably be labeled, and you will have to identifywhich point has the greatest or least velocity. Remember, the pointwith the greatest slope has the greatest velocity, and the point withthe least slope has the least velocity. The turning points of thegraph, the tops of the “hills” and the bottoms of the “valleys” wherethe slope is zero, have zero velocity.
[align=center][img]http://www.24en.com/d/file/sat/sat2/physics/2008-01-24/ce2887501dbe1582aab8eb24b74cf887.gif[/img][/align] In this graph, for example, the velocity is zero at points [i]A[/i] and [i]C[/i], greatest at point [i]D[/i], and smallest at point [i]B[/i]. The velocity at point [i]B [/i]is smallest because the slope at that point is negative. Because velocity is a vector quantity, the velocity at [i]B[/i] would be a large negative number. However, the speed at [i]B [/i]is greater even than the speed at [i]D[/i]: speed is a scalar quantity, and so it is always positive. The slope at [i]B [/i]is even steeper than at [i]D[/i], so the speed is greatest at [i]B[/i].
[b] Velocity vs. Time Graphs[/b]
Velocity vs. time graphs are the mosteloquent kind of graph we’ll be looking at here. They tell us verydirectly what the velocity of an object is at any given time, and theyprovide subtle means for determining both the position and accelerationof the same object over time. The “object” whose velocity is graphedbelow is our ever-industrious ant, a little later in the day.
[align=center][img]http://www.24en.com/d/file/sat/sat2/physics/2008-01-24/8c4f2288af618a47e0f8c44d205270d3.gif[/img][/align]
We can learn two things about the ant’svelocity by a quick glance at the graph. First, we can tell exactly howfast it is going at any given time. For instance, we can see that, twoseconds after it started to move, the ant is moving at 2 cm/s. Second, we can tell in which direction the ant is moving. From [i]t [/i]= 0 to [i]t[/i] = 4, the velocity is positive, meaning that the ant is moving to the right. From [i]t[/i] = 4 to [i]t[/i] = 7, the velocity is negative, meaning that the ant is moving to the left.
[b] Calculating Acceleration[/b]
We can calculate acceleration on a velocityvs. time graph in the same way that we calculate velocity on a positionvs. time graph. Acceleration is the rate of change of the velocityvector, [img]http://www.24en.com/d/file/sat/sat2/physics/2008-01-24/ec4223838f40e822cacd9ab8e87be58e.gif[/img], which expresses itself as the slope of the velocity vs. time graph. For a velocity vs. time graph, [i]the acceleration at time [/i]t[i] is equal to the slope of the line at [/i]t[i].[/i]
What is the acceleration of our ant at [i]t[/i] = 2.5 and [i]t[/i] = 4? Looking quickly at the graph, we see that the slope of the line at [i]t[/i] = 2.5 is zero and hence the acceleration is likewise zero. The slope of the graph between [i]t[/i] = 3 and [i]t[/i] = 5 is constant, so we can calculate the acceleration at [i]t[/i] = 4 by calculating the average acceleration between [i]t[/i] = 3 and [i]t[/i] = 5:
[align=center][img]http://www.24en.com/d/file/sat/sat2/physics/2008-01-24/103bf31ea8e7fe54148962adff2f4fb5.gif[/img][/align] The minus sign tells us that acceleration is in the leftward direction, since we’ve defined the [i]y[/i]-coordinates in such a way that right is positive and left is negative. At [i]t[/i] = 3, the ant is moving to the right at 2cm/s, so a leftward acceleration means that the ant begins to slowdown. Looking at the graph, we can see that the ant comes to a stop at [i]t[/i] = 4, and then begins accelerating to the right.
[b] Calculating Displacement[/b]
Velocity vs. time graphs can also tell usabout an object’s displacement. Because velocity is a measure ofdisplacement over time, we can infer that:
[align=center][img]http://www.24en.com/d/file/sat/sat2/physics/2008-01-24/d0dc23b8f22bfe3700738340dd5639e0.gif[/img][/align]
Graphically, this means that [i]the displacement in a given time interval is equal to the area under the graph during that same time interval.[/i] If the graph is above the [i]t[/i]-axis, then the positive displacement is the area between the graph and the [i]t[/i]-axis. If the graph is below the [i]t[/i]-axis, then the displacement is negative, and is the area between the graph and the [i]t[/i]-axis. Let’s look at two examples to make this rule clearer.
First, what is the ant’s displacement between [i]t[/i] = 2 and [i]t[/i] = 3? Because the velocity is constant during this time interval, the area between the graph and the [i]t[/i]-axis is a rectangle of width 1 and height 2.
[align=center][img]http://www.24en.com/d/file/sat/sat2/physics/2008-01-24/6110c7672ea9698dca3b78ad32d23082.gif[/img][/align]
The displacement between [i]t[/i] = 2 and [i]t[/i] = 3 is the area of this rectangle, which is 1 cm/s[img]http://www.24en.com/d/file/sat/sat2/physics/2008-01-24/912b4b321e1ea83ca44f2f5c0e539169.gif[/img]s = 2 cm to the right.
Next, consider the ant’s displacement between [i]t[/i] = 3 and [i]t[/i] = 5. This portion of the graph gives us two triangles, one above the [i]t[/i]-axis and one below the [i]t[/i]-axis.
[align=center][img]http://www.24en.com/d/file/sat/sat2/physics/2008-01-24/aea123a7cce4d05f3bca5975744c64f7.gif[/img][/align]
Both triangles have an area of 1 /2(1 s)(2 cm/s) = 1 cm. However, the first triangle is above the [i]t[/i]-axis, meaning that displacement is positive, and hence to the right, while the second triangle is below the [i]t[/i]-axis, meaning that displacement is negative, and hence to the left. The total displacement between [i]t[/i] = 3 and [i]t[/i] = 5 is:
[align=center][img]http://www.24en.com/d/file/sat/sat2/physics/2008-01-24/7c34f3e209b67401a48e8ef1fc51feaf.gif[/img][/align]
In other words, at [i]t[/i] = 5, the ant is in the same place as it was at [i]t[/i] = 3.
[b] Curved Velocity vs. Time Graphs[/b]
As with position vs. time graphs, velocityvs. time graphs may also be curved. Remember that regions with a steepslope indicate rapid acceleration or deceleration, regions with agentle slope indicate small acceleration or deceleration, and theturning points have zero acceleration.
[b] Acceleration vs. Time Graphs[/b]
After looking at position vs. time graphsand velocity vs. time graphs, acceleration vs. time graphs should notbe threatening. Let’s look at the acceleration of our ant at anotherpoint in its dizzy day.
[align=center][img]http://www.24en.com/d/file/sat/sat2/physics/2008-01-24/6249e84a1aa18a22fd1222402692c60f.gif[/img][/align]
Acceleration vs. time graphs give usinformation about acceleration and about velocity. SAT II Physicsgenerally sticks to problems that involve a constant acceleration. Inthis graph, the ant is accelerating at 1 m/s2 from [i]t[/i] = 2 to [i]t[/i] = 5 and is not accelerating between [i]t[/i] = 6 and [i]t[/i] = 7; that is, between [i]t[/i] = 6 and [i]t[/i] = 7 the ant’s velocity is constant.
[b] Calculating Change in Velocity[/b]
Acceleration vs. time graphs tell us aboutan object’s velocity in the same way that velocity vs. time graphs tellus about an object’s displacement. [i]The change in velocity in a given time interval is equal to the area under the graph during that same time interval.[/i] Be careful: the area between the graph and the [i]t[/i]-axis gives the [i]change[/i] in velocity, not the final velocity or average velocity over a given time period.
What is the ant’s change in velocity between [i]t[/i] = 2 and [i]t[/i] = 5? Because the acceleration is constant during this time interval, the area between the graph and the [i]t[/i]-axis is a rectangle of height 1 and length 3.
[align=center][img]http://www.24en.com/d/file/sat/sat2/physics/2008-01-24/36b35378c7aa6ba33b79d51d710201a8.gif[/img][/align]
The area of the shaded region, and consequently the change in velocity during this time interval, is 1 cm/s2 · 3 s = 3 cm/s to the right. This doesn’t mean that the velocity at [i]t[/i] = 5 is 3 cm/s; it simply means that the velocity is 3 cm/s greater than it was at [i]t[/i] = 2. Since we have not been given the velocity at [i]t[/i] = 2, we can’t immediately say what the velocity is at [i]t[/i] = 5.
[b] Summary of Rules for Reading Graphs[/b]
You may have trouble recalling when to lookfor the slope and when to look for the area under the graph. Here are acouple handy rules of thumb:
[list=1][*]The slope on a given graph is equivalent to the quantity we get by dividing the [i]y[/i]-axis by the [i]x[/i]-axis. For instance, the [i]y[/i]-axis of a position vs. time graph gives us displacement, and the [i]x[/i]-axisgives us time. Displacement divided by time gives us velocity, which iswhat the slope of a position vs. time graph represents.[*]The area under a given graph is equivalent to the quantity we get by multiplying the [i]x[/i]-axis and the [i]y[/i]-axis. For instance, the [i]y[/i]-axis of an acceleration vs. time graph gives us acceleration, and the [i]x[/i]-axisgives us time. Acceleration multiplied by time gives us the change invelocity, which is what the area between the graph and the [i]x[/i]-axis represents.[/list]
We can summarize what we know about graphs in a table:
[align=center][img]http://www.24en.com/d/file/sat/sat2/physics/2008-01-24/951248e6fd81357d9bc0e572a15f6485.gif[/img][/align]
