Motion in Two Dimensions Study Guide
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.
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