Motion in Two Dimensions Study Guide (page 3)
In this lesson, we will continue the study of important motion quantities and learn about motion in two dimensions, relative velocity, and projectile motion.
A vector can represent behavior over three separate directions x, y, and z. The same concepts of displacement, average and instantaneous velocity, and average and instantaneous acceleration will apply to motion in two dimensions. The major difference is that we will work with formulas containing only one variable at a time: x or y or z. This is a real advantage, as in separate directions, different interactions might predominate, and therefore, the source of interaction (and the specific force) will be considered for separate directions.
In this case, the vector position for each of the initial and final positions is determined by two coordinates: x and y.
Δ x = xf – xi
Δ y = yf – yi
And from these components, the displacement of the vector is:
Δ r = (Δ x, Δ y)
Δ r = Δ x + Δ y
The displacement vector is the segment that connects the initial and final positions, and is directed toward the final position:
Δ r = rf – bi
A toy car starts from the corner of the classroom and travels to a new position of coordinates (5,6) meters away from the corner. Draw the displacement and find the magnitude.
We can see that the two measurements in the problems are expressed in meters, so no conversion is necessary. We will draw a system of coordinates, and we will place the car in the origin of the system at the beginning of motion. This is the initial position. The final position is given by the problem. We will construct the displacement as the vector that connects the initial and final position and then find the magnitude by using the components.
Now the problem becomes a one-dimensional problem because we can write the equations for motion on the separate axis (see Figure 4.1):
(xi, yi) = (0,0) m
(xf, yf ) = (5,6) m
Δ x = xf – xi
Δ x = 5 – 0 = 5 m
Δ y = yf – yi
Δ y = 6 – 0 = 6 m
Δ r = (5,6) m
Δ r = (52 + 62)1/2 = 8 m
If the motion is more complicated (for example, a nonlinear trajectory), then we would have cases where the distance traveled is not necessarily the same as the displacement. For example, in the case shown in Figure 4.2, the distance is larger than the displacement.
A car travels 40 miles to the west and then another 30 miles to the north. Draw the displacement, calculate the magnitude and direction of the displacement, and compare to the distance traveled.
As the single values of this problem are in miles and no other requirement exists for conversion in the problem, we can keep the solution in miles. If we are asked to convert to SI, we would multiply each of the two distances with the conversion factor 1,609 meters/1 mile to determine the result in meters.
Δ ri = 40 miles W
Δ rf = 30 miles N
Δ rtotal = ?
distance = ?
The distance traveled by the car is the sum of the two distances:
Distance = 40 miles + 30 miles = 70 miles
The total displacement has to take into account each of the displacements, but we cannot simply add the two numbers because their vector expressions put them in different directions. Let us draw a motion diagram. We will consider east as the positive x-axis and north as the positive y-axis (see Figure 4.3).
(Δ rtotal)2 = (Δ ri )2 + (Δ rf )2
Δ rtotal = [(Δ ri)2 + (Δ rf)2]1/2
Δ rtotal = 50 miles
The direction is the angle θ, and it can be calculated from the geometry of the problem:
sin θ = rf /rtotal = 30 miles/50 miles
θ = sin–1 (0.6) = 37°
Hence, displacement is Δ rtotal = 50 miles at 37° north of west or (40 miles,30 miles) by the x and y components.
It can be seen that the distance is larger compared to the displacement, as shown in the figure (the two sides of the triangle added together are larger than the hypotenuse).
Based on the definition of displacement, we introduce also the average velocity.
When the time interval is very small (approaching zero, or Δ t→ 0), the definition of the average velocity becomes the definition of the instantaneous velocity.
If the motion is such that velocity is constant (both in magnitude and direction), then the average speed and the instantaneous velocities are the same.
Average velocity is the time rate of change of displacement:
A toy car starts moving at a constant speed for 10 seconds. Show with vectors, at two separate positions, the average and instantaneous velocities.
We will represent the car in a two-dimensional motion by using a point-like object and drawing the two velocities. They are equal vectors (this means that both direction and magnitude are the same). See Figure 4.4.
When the trajectory is nonlinear, instantaneous velocity is a vector always tangent to the trajectory as shown in Figure 4.5.
Instantaneous velocity is velocity at a certain time:
Use a diagram to show the displacement of a car that moves between the following positions: (50,30) meters to (–40,40) meters in 4 seconds, and then calculate displacement and the average velocity, both expressed with their coordinates and magnitude and angle.
All units are expressed in SI. In order to calculate the average velocity, we need to determine displacement. Figure 4.6 shows the initial and final positions.
Δx = xf – xi
Δx = – 40 – 50 = – 90 m
Δy = yf – yi
Δy = 40 – 30 = 10 m
Δr = (–90,10) m
Δr = (902 + 102)1/2 = 91 m
tan θ = Δy/Δx = 10 m/( –90 m) = –1/9
θ = tan–1(1/9) = –6°
Therefore, displacement is 91 meters at an angle of 6° north of west.
Average velocity is:
Vaverage = (22.52 + 2.52)1/2 = 23 m/s
θ = tan–1 (–2.5/22.5) = –6°
Or average velocity is 23 m/s (also = 91 m/4 s) at an angle of 6° north of west.
We can see that the average velocity and displacement are in the same direction. As noted previously, the instantaneous velocity will be tangent to the trajectory at every point but not necessarily in the same direction as displacement and average velocity.
If the motion is made with changing velocity, then acceleration can be defined.
Similar to the average and instantaneous velocity, starting with the definition just given, when the time interval approaches zero, one can define instantaneous acceleration.
As seen in both these definitions, the vector acceleration has the direction of the change in velocity, and that can be determined by either change in magnitude or change in direction.
Average acceleration is the time rate of change of velocity:
Consider an object moving on a curve as shown in Figure 4.7. Draw the vector average acceleration based on the information in the graph.
Start by constructing the vector Δv and then draw the average acceleration aaverage based on this direction.
Because average acceleration is nothing else, then by dividing a vector (Δv) by a scalar (Δt), there will be no change in direction. Only magnitude will be increased or decreased, depending on the size of the denominator.
Instantaneous acceleration is acceleration at a certain time:
As you have probably started to notice, a lot in mechanics is relative. All measurements start with distance and time, neither of which have an absolute origin that can be easily applied to all sorts of motion. So, what solution do we have for this issue? The answer is to consider our own reference frames and origins of space and time. You have considered a simple, one-axis system of reference in linear motion. In this chapter, we worked on a reference system composed of two perpendicular axes: x and y. In real space, a need exists for a third axis—height, or the z-axis. And even more, some consider that time be represented by a fourth coordinate, because motion spans not only in space but also in time. Depending on the axes of reference, motion looks different. Imagine yourself in space, far away from any cosmic object. Is your ship moving or not? If the engines are quiet, and there is no acceleration, you will not be able to tell your motion if you are not looking at your senso. But if an alien species is watching you with deep space sensors from a far away planet, they will be able to tell if you are moving and how are you moving. So, who is right? Sure enough, the answer is both observants: the alien species and you. The difference is that each of you relates to different frames: Your frame moves with you, so evidently, you cannot determine whether you are at rest or not, while the aliens analyze your motion relative to a different system and see that you are in motion with respect to their system. This is why for more down-to-earth motion, we define relative velocity.
The fixed frame reference for most of our problems will be Earth. We will not consider Earth's motion around its own axis or around the sun when we are considering an apple falling from a tree or a car or boat traveling for some distance. In other cases, such as for space travel, a distant star can be considered as a fixed position. Although this is not completely true (the distant star is moving), if the stellar object is far enough away, the motion during the experiment might be negligible compared to the motion of the analyzed system.
Relative velocity is the resultant vector measuring the velocity of an object in a fixed frame.
vr = v1 – v2
Where v1 is the speed of an object with respect to a moving frame and v2 is the speed of the frame with respect to a fixed frame.
While biking, you keep your velocity at a constant 12 m/s magnitude directed east with respect to the earth. The wind starts blowing toward the west with a speed of 5 m/s, with respect to the earth. What is your resultant speed with respect to the earth at the time the wind starts blowing.
We will draw a diagram to show the two velocities. The resultant velocity is the vector sum of the two velocities considered in the problem. We will consider east to be the positive x-axis direction (see Figure 4.8).
Hence, the resultant speed is:
vr = v1 – v2
vr = v1 – v2
vr = 12 – 5 = 7 m/s
vr is 7 m/s in the east direction.
A classic case of two-dimensional motion is projectile motion. Imagine you throw an object horizontally, as in skipping a rock. In a gravity-free and friction-free world, the rock will move horizontally until it interacts with an obstacle as shown in Figure 4.9. The diagram shows the constant velocity (magnitude and direction).
If we consider the friction with the air, the rock will use energy through conversion to heat during its path and it will slow down. Figure 4.10 shows the velocity and the acceleration (constant acceleration).
If we consider no friction but an acceleration downward, as in the case of the acceleration due to gravity, g = 9.8 m/s2, what happens to the velocity? You might surmise the answer from everyday happenings, but this is a good place to use what we learned and to learn something new. The graph in Figure 4.11 shows the object moving down, and then acceleration due to gravity is also down. As we have seen before, the acceleration is in the same direction as the change in velocity, hence the change in velocity, Δv, has to be also downward. But Δv = v2 – v1 and that means according to the diagram that v2 has to be the same direction as v1 and larger than v1.
Hence, we will complete the diagram in Figure 4.11 with the vector velocity at other positions (see Figure 4.12).
What will happen if the object moves up? Then the acceleration is opposite to the initial velocity, and we have a case similar to the previous one with friction. The object will slow down.
In the previous discussion, we separated the horizontal and vertical motion, and we examined different motion. What happens if we consider a more complex motion on both x and y direction? We will be able then to add the x and y components of displacement, velocity, and acceleration and then fully characterize the two-dimensional motion: We obtain projectile motion.
A rock is thrown with a speed of 10 m/s at an angle of 30 degrees from the horizontal. Draw a diagram of motion and show in a few places, both when the rock is ascending and descending, the velocity, and the acceleration.
v0 = 10 m/s
α = 30°
We will consider a two-dimensional frame and a friction-free motion.
On the horizontal x-axis, the object moves with constant velocity (vx = constant and vx = 0 m/s2) because there is no friction and no other interaction. Also, if the object does not start from the origin of the reference frame, an offset has to be considered, x0:
x = x0 + vx · Δ t
and vx is the component of the speed on the x-axis (see Figure 4.13).
In the vertical direction, the object moves up decelerating (slowing down), stops at the top of the trajectory, and then returns down accelerating (see Figure 4.14). All this motion is determined by the constant acceleration due to gravity.
The initial components of the velocity can be found from the data in Figure 4.15.
v0x = v0 · cos α
v0y = v0 · sin α
With all this information, the x-y graphical description of the previous motion is shown in Figure 4.16.
The maximum height (vertical) and distance (horizontal) the object reaches is based on the initial speed and the angle of throw. The horizontal distance for a symmetric projectile motion as in this previous example (the projectile starts and finishes at the same y position) is called range. To summarize the equation of motion for the x- and y-axis are:
v0x = v0 · cos α = constant
x = x0 + v0 · cos α Δt
ax = 0 m/s2
v0y = v0 · sin α
y = y0 + v0y · Δt + ay · Δt2/2
ay = constant
If we eliminate the time dependence from both x(t) and y(t), then we get the y(x) dependence:
And we can determine the range and the maximum height to be:
In these equations, the initial position is considered to be the origin:
x0 = 0
y0 = 0
The time it takes the object to describe this trajectory is called time of flight and can be calculated to be:
Figure 4.17 shows the different paths an object takes for a fixed speed with different throwing angles. The longest path (maximum range at a given speed) is for the 45°.
Practice problems of this concept can be found at: Motion in Two Dimensions Practice Questions
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