Chapter Review
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Chapter 3: Vectors in Physics Chapter Review |
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3-5 Position, Displacement, Velocity, and Acceleration Vectors
If you recall the discussions of position, displacement, velocity, and acceleration from chapter 2 you may remember that associated with each of these quantities was a size (magnitude) and a direction (positive or negative). Therefore, these are all vector quantities and in more than one dimension we wish to represent these quantities in the more general vector notation discussed above. The SI units of all these quantities is stated in chapter 2.
The position vector r for an object in a coordinate system is the arrow from the origin of the coordinate system to the location of the object. The magnitude of this vector is the distance of the object from the origin. The x and y scalar components of r are the x and y coordinates of the object. Therefore, in unit vector notation the two-dimensional position vector is written as
r =x + y
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Examples 3.1 and 3.2 involved two-dimensional position vectors.
The displacement of an object is its change in position. So the displacement vector Dr is the difference between its final and initial positions Dr = rf - ri. Given what we learned in the section 3-3 about vector addition and subtraction, the unit vector notation for displacement in two dimensions is
Dr = (xf - xi)
The average velocity is the displacement divided by the elapsed time Dt. Dividing the vector Dr by the scalar Dt, we divide each component of Dr by Dt. Hence,
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where Dx = xf - xi and Dy = yf - yi. The instantaneous velocity, once again, is given by the limit of the average velocity as the time interval approaches zero
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The average acceleration is the change in velocity Dv divided by the elapsed time Dt
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The change in velocity is the difference between the final and initial velocities,
Dv = vf - vi. The instantaneous acceleration is given by the limit of the average acceleration as the time interval approaches zero
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Example 3.6 Out of Gas: Suppose you're driving northwest at 13.5 m/s for exactly one half-hour when you run out of gas. Frustrated, you walk for 40.0 minutes to the nearest gas station which is 2.40 km away at 30o north of west from where your car stopped. Determine your average velocity for this trip. State the answer both as a magnitude and a direction and using unit vectors.
Picture the Problem The picture shows the three relevant displacement vectors for the trip. The vector dc is your displacement while driving the car, dw is your displacement while walking, and r is your total displacement.
Strategy Since average velocity is displacement divided by elapsed time, we will perform the calculations here by first calculating the displacement vectors then determining the average velocity.
Solution
Step 1: determine the displacement dc:
| 1. Calculate the magnitude of dc from the speed and time while in the car: |  | 2. Going northwest implies that the direction of dc is: |  |
| 3. Now determine the x- and y-components of dc: |   |
Step 2: determine the displacement dw:
| 1. 30o north of west implies that the direction of dw is: |  |
| 2. Now determine the x- and y-components of dw: |   |
Step 3: The total displacement r is
dc + dw:
| 1. Determine the x- and y-components of r: | rx = dcx + dwx = -17,183 m - 2,078 m = -19,261 m ry = dcy + dwy = 17,183 m + 1,200 m = 18,383 m |
| 2. Determine the total elapsed time t: |  |
| 3. Determine the x- and y-components of vav: |  |
| 4. State the average velocity using unit vectors: | vav = -(4.59 m/s) + (4.38 m/s) |
| 5. Find the magnitude of vav: |  |
6. Calculate a reference angle  for vav: |
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| 7. From the signs of the components we can see that is measured above the -x axis. Therefore, the magnitude and direction of vav can be stated as: |
6.34 m/s at 43.7o north of west. |
Insight In the final calculation of q the negative signs comes strictly as a result of the negative x-component. Without any further statement it would be assumed that the angle is measured clockwise from the +x-axis. This potential mistake is why the interpretation in step 7 is needed.
Exercise 3.7 Turn left at the light: Suppose that you are driving at 35 mi/h down Main St. while approaching a green light at the intersection with State Street. In order to turn left onto State St. you slow down, make the turn, then speed up to 35 mi/h on State Street. If it requires a total time of 1.75 minutes from the time you begin to slow down on Main St. until you reach 35 mi/h on State St., what is the magnitude and direction of your average acceleration?
Solution Try to sketch the velocity and average acceleration vectors for this problem. Let us choose the direction of the initial velocity on Main St. to be the +x direction and the direction of the final velocity on State St. to be the +y direction.
Given: vi = 35 mi/h,
Find: aav,
Since the average acceleration is determined by the change in velocity Dv, let us first calculate Dv by getting the components of the final and initial velocities. For the initial velocity we have
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For the final velocity we have
Now that we have the components we can determine the components of Dv.
From these components we can determine the magnitude.
The total elapsed time is 1.75 x 60 s = 105 s. So the magnitude of the average acceleration is
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To determine the direction, notice that aav,x would be negative, because Dvx is negative, while aav,y would be positive for a similar reason. These facts mean that aav lies in the second quadrant of the coordinate system. (If you couldn't sketch aav above then do so now.) Also notice that since the x- and y-components of Dv have the same size, 15.6 m/s, then Dv, and therefore aav must make a 45 degree angle in that quadrant. So finally, we can say that the average acceleration is 0.21 m/s2 at 135 degrees from the positive x-direction.
Notice that even though the magnitudes of the initial and final velocities are the same, the acceleration is not zero. This is because the directions of the velocities are different and this difference is just as important for determining acceleration as a difference in speed.
Physlet Illustration: Velocity and Acceleration Vectors | |
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The blue ball moves in two dimensions. Can you determine
the directions of its velocity and acceleration vectors at any instant? Start |
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Hints
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Practice Quiz
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 - 11.9 |
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 - 1.6 |
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+ 11.5 |
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3-6 Relative Motion
A good and relevant application of vector addition and subtraction is relative motion. Specifically, we will focus on the velocity of an object as measured by two observers who are moving with respect to each other at a constant relative velocity. To identify who is measuring what, we use a system of subscripts in which the first subscript refers to the object whose velocity is being measured and the second subscript refers to the observer (or coordinate system) making the measurement. [Note that the observer can be fictitious; we really only need the coordinate system with respect to which the velocity has the value in question]. For example, a velocity labeled as v12 refers to the velocity of object 1 relative to (or as measured by) object 2. If you reverse the subscripts to get the velocity of 2 relative to 1 this velocity v21 relates to v12 by a minus sign: v21 = -v12.
To state how relative velocity motion is handled in more general terms, consider the diagram below which shows two coordinate systems A and B, and a point P that is in motion relative to each coordinate system. Observers in system A identifies the velocity of P as vPA; while observers in system B identifies P's velocity as vPB. Let us assume that system B moves relative to system A with a velocity vBA. The relationship between these velocities can be written as
vPB = vPA + vBA.
It is very helpful to notice how the subscript convention works in this velocity addition equation. The resultant velocity vPA has P as leftmost subscript and A as rightmost subscript. Now, the subscripts on the two velocities whose addition produces the resultant are ordered PBBA. In this ordering P is the leftmost subscript and A is the rightmost just as with the resultant. The subscript for the system not referenced by the resultant (B in this case) is repeated in the middle. This mnemonic can be used to help analyze almost any relative velocity situation as long as you stay consistant with how you label the velocities. You should be careful, however, not to fall into the trap of using this mnemonic as a substitute for understanding the physical situation being described.
Example 3.8 Crossing the Street: Tom and Jan are standing at the side of the street waiting to cross. A car is coming at 10.0 m/s. Jan decides to go ahead and run directly across the street at 2.80 m/s while Tom chooses to wait. While jan is running across, what is her velocity as measured by the driver of the car?
Picture the Problem The picture indicates the motion of both Jan and the car. The velocity vectors are labled such that "g" means ground.
Strategy We need to determine Jan's velocity with respect to the car. In our system of labels this velocity is vJc.
Solution
| 1. The vector addition we need is: | vJc = vJg + vgc |
| 2. Write out vJg based on given information: | vJg = 2.80 m/s |
| 3. Write out vgc based on given information: | vgc = -vcg = -10.0 m/s |
| 4. Performing the vector addition yields: | vJc = -(10.0 m/s) + (2.80 m/s) |
Insight The mnemonic outlined above made it straightforward to identify the correct vector addition to solve this problem. However, make sure that the final answer for vJc makes intuitive sense to you as well. Why should its x-component be negative?
Relative Velocity | |
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| Start | |
| Driving peacefully along the road in your new VW Beetle, you are suddenly passed by a red Ferrari. What is the relative velocity of the Ferrari with respect to you? The clock time shown is in seconds and grid distances are in meters. (Positive is to the right, and negative is to the left.) | |
Hints
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Example 3.9 Holding an Umbrella: You are walking on campus at 1.3 m/s directly against a 20 mi/h, horizontal wind in a light snow storm. If the snowflakes fall vertically downward with a speed of 3.9 m/s with respect to still air, at what angle, with respect to the vertical, should you hold your umbrella to best protect yourself from the snow?
Picture the Problem The picture shows you, relative to the ground, walking in the storm holding the umbrella at an angle. The motion of the snow relative to the ground is also indicated by the lines about to impact the umbrella.
Strategy The umbrella protects best if it is oriented so that the snow impacts the greatest amount of its surface. For this to happen the shaft of the umbrella should be oriented along the line of motion of the snow (as viewed by you). Therefore, the question will be answered if we determine the angle that the velocity of the snow makes with the vertical as measured by you; so, we need vsy.
Solution
The various reference frames here are you (y), the ground (g), the snow (s), and the air (a).
| 1. The vector addition that we need is: | vsy = vsg + vgy |
| 2. We can immediately write down vgy: | vgy = -vyg = -1.3 m/s |
| 3. The vector addition to get vsg is: | vsg = vsa + vag |
| 4. vsa is given in the problem: | vsa = -3.9 m/s |
| 5. From the given information we can also write vag: |  |
| 6. The velocity of the snow relative to the ground is: | vsg = -8.94 m/s - 3.9 m/s |
| 7. Using step 1 the velocity of the snow relative to you becomes: |  |
| 8. As shown in the sketch below, the angle D this velocity makes with the vertical is: |  |
| 9. As discussed in the strategy section, this must also be the angle at which you should hold the umbrella. |  |
Insight It would have been easy to forget that the angle of the umbrella is determined relative to you. A common mistake here would have been to only calculate the angle of snow relative to the ground. When dealing with relative motion always remember to ask yourself with respect to whom is this question being asked.
Practice Quiz
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- 30 |
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