describing motion with diagrams
read from lesson 2 of the 1-d kinematics chapter at the physics classroom:
http://www.physicsclassroom.com/class/1dkin/u1l2a.html
http://www.physicsclassroom.com/class/1dkin/u1l2b.html
http://www.physicsclassroom.com/class/1dkin/u1l2c.html
mop connection: kinematic concepts: sublevel 5
motion can be described using words, diagrams, numerical information, equations, and graphs. using diagrams to describe the motion of objects involves depicting the location or position of an object at regular time intervals.
1. motion diagrams for an amusement park ride are shown. the diagrams indicate the positions of the car at regular time intervals. for each of these diagrams, indicate whether the car is accelerating or moving with constant velocity. if accelerating, indicate the direction (right or left) of acceleration. support your answer with reasoning.
table with diagrams a - e, columns: acceleration (y/n, dirn), reason blanks
2. suppose that in diagram d (above) the cars were moving leftward (and traveling backwards). what would be the direction of the acceleration? ______ explain your answer fully.

Question

Accepted Answer

### Part 1: Analyzing Each Motion Diagram

#### Diagram a
# Brief Explanations:
To determine acceleration, we check the distance between consecutive positions (at regular time intervals). If the distance is constant, velocity is constant (no acceleration). In diagram a, the spacing between the car’s positions is uniform. So, constant velocity means no acceleration.
# Answer:
- Reason: Equal spacing (constant velocity).
- Acceleration: N (No), Dir’n: (Blank)

#### Diagram b
# Brief Explanations:
Acceleration occurs when velocity changes (either speed or direction). Here, the spacing between the car’s positions **increases** as it moves right. Larger spacing means increasing speed (positive acceleration) in the direction of motion (right).
# Answer:
- Reason: Spacing increases (speeding up right).
- Acceleration: Y (Yes), Dir’n: Right

#### Diagram c
# Brief Explanations:
Uniform spacing between positions means the car covers equal distance in equal time intervals. Thus, velocity is constant, so no acceleration.
# Answer:
- Reason: Equal spacing (constant velocity).
- Acceleration: N (No), Dir’n: (Blank)

#### Diagram d
# Brief Explanations:
The spacing between the car’s positions **increases** as it moves right. Increasing spacing implies the car is speeding up (acceleration) in the direction of motion (right).
# Answer:
- Reason: Spacing increases (speeding up right).
- Acceleration: Y (Yes), Dir’n: Right

#### Diagram e
# Brief Explanations:
The spacing between the car’s positions **decreases** as it moves right (or increases as it moves left, depending on interpretation). A decreasing spacing in the direction of motion (right) means the car is slowing down, but the “Dir’n” here refers to acceleration direction. Wait—actually, if spacing decreases while moving right, the car is decelerating (negative acceleration) relative to right, but the diagram shows spacing getting smaller toward the right (so speed is decreasing in the right direction, meaning acceleration is left). Alternatively, if we see the car “bunching up” toward the right, it’s slowing down while moving right, so acceleration is left.
# Answer:
- Reason: Spacing decreases (slowing down right, so acceleration left).
- Acceleration: Y (Yes), Dir’n: Left

### Part 2: Diagram D (Leftward Motion)
# Brief Explanations:
In diagram D, originally, the car moves right with increasing spacing (speeding up right). If it moves leftward (opposite direction) with the same spacing pattern (increasing spacing as it moves left), this means:
- Direction of motion: Left.
- Spacing between positions: Increases as it moves left (so it’s speeding up in the left direction? Wait, no—wait, the original diagram D has spacing increasing to the right. If motion is leftward, the “spacing” between positions (from left to right) would still be increasing, but the car is moving left. So the distance between consecutive positions (as time increases) is increasing, but the car is moving left. So velocity is leftward, and speed is increasing (because spacing between positions is increasing). Acceleration direction is the same as the direction of increasing speed. If speed is increasing while moving left, acceleration is left? Wait, no—let’s clarify:

Velocity is a vector (direction + speed). If the car moves left and its speed (magnitude of velocity) is increasing, then acceleration is in the direction of motion (left). But wait, in the original diagram D, the car moves right and speeds up (acceleration right). If we reverse the motion (leftward), the spacing between positions (from left to right) is still increasing, meaning the car covers more distance (in the leftward direction) each time interval. So speed (magnitude) is increasing in the left direction. Thus, acceleration is left? Wait, no—wait, acceleration is the rate of change of velocity. If velocity is leftward (negative, say) and speed is increasing (becoming more negative), then acceleration is leftward (same direction as velocity).

Alternatively, think of “speeding up” vs. “slowing down”:
- If velocity and acceleration are in the same direction: speeding up.
- If opposite: slowing down.

In the original D, motion is right (velocity right), acceleration right (speeding up right).

If motion is left (velocity left), and the spacing between positions (as time increases) is increasing (meaning the car moves left faster each interval), then velocity (left) and acceleration (left) are in the same direction: speeding up left. Thus, acceleration direction is left? Wait, no—wait, the spacing in diagram D (original) is: first two cars are close, then spacing increases. So time 1: position 1; time 2: position 2 (closer to position 1); time 3: position 3 (farther from position 2); etc. Wait, no—original diagram D: positions are at times t1, t2, t3, t4, t5 (from left to right). The spacing between t1-t2 is small, t2-t3 is larger, t3-t4 larger, t4-t5 larger. So as time increases (moving right), spacing increases: speeding up right.

If motion is leftward, the car moves from t5 to t4 to t3 to t2 to t1 (leftward). So at t5 (rightmost), then t4 (left of t5), t3 (left of t4), etc. The spacing between t5-t4 is small, t4-t3 is larger, t3-t2 larger, t2-t1 larger. So as time increases (moving left), the distance covered (leftward) increases each interval: speed is increasing in the left direction. Thus, velocity is left, acceleration is left (same direction: speeding up left). Wait, but that seems contradictory. Wait, no—if you move left and your speed (how fast you move left) is increasing, acceleration is left. If you move left and speed decreases, acceleration is right.

But in the original diagram D, the car is speeding up right (acceleration right). If we reverse the motion (leftward), the “speeding up” would be in the left direction, so acceleration left? Wait, no—let’s use the definition: acceleration is the change in velocity over time. Velocity is (displacement)/(time). If the car moves left, displacement is negative (if right is positive). The change in velocity: if spacing increases (distance between positions increases) as it moves left, then each time interval, the displacement (leftward) is larger. So velocity at t2 (leftward) is more negative than at t1, velocity at t3 is more negative than at t2, etc. So the change in velocity (Δv) is negative (more negative) over positive time, so acceleration (Δv/Δt) is negative (leftward). Thus, acceleration direction is left.

Wait, but maybe a simpler way: in the original diagram, the car is speeding up in the right direction (acceleration right). If we make the car move left, the “speeding up” would be in the left direction, so acceleration left. But wait, no—if the spacing between positions (when moving left) is increasing, that means the car is moving left faster each time, so acceleration is left.

But let’s check the original diagram D: the spacing between the cars increases as you go right. So from left to right, the cars are getting farther apart. So time 1: leftmost car; time 2: next car (right of time 1, closer); time 3: next (farther from time 2); etc. So the car moves right, and the distance between consecutive positions (time intervals) increases: speeding up right (acceleration right).

If the car moves left, it’s moving from the rightmost car to the leftmost, so time 1: rightmost; time 2: left of time 1 (closer to time 1); time 3: left of time 2 (farther from time 2); etc. So the distance between time 1 and time 2 is small (car moves left a little), time 2 to time 3 is larger (car moves left more), etc. So the car’s speed (magnitude of velocity) is increasing as it moves left. Thus, velocity is left, acceleration is left (same direction: speeding up left). Wait, but that would mean acceleration is left. But let’s think again: if you move left and your speed is increasing, acceleration is left. If you move left and speed decreases, acceleration is right.

But in the original diagram, the car is speeding up right (acceleration right). If we reverse the motion (leftward), the “speeding up” is in the left direction, so acceleration left. However, another way: the spacing between the cars (in the diagram) is increasing to the right. So the pattern is: small, medium, large, larger. If the car moves left, the order of positions is reversed: larger, large, medium, small. So the distance between consecutive positions (as time increases) is decreasing (from larger to large to medium to small). Wait, no—time increases from left to right in the diagram. So time 1: leftmost car; time 2: next (right of time 1); time 3: next (right of time 2); etc. So when moving left, the car goes from time 5 (rightmost) to time 4 to time 3 to time 2 to time 1 (leftmost). So the distance between time 5 and time 4 is the spacing between the rightmost two cars (small), time 4 to time 3 is the next spacing (medium), time 3 to time 2 (large), time 2 to time 1 (larger). Wait, no—original diagram D: the cars are at positions with spacing: t1-t2 (small), t2-t3 (medium), t3-t4 (large), t4-t5 (larger). So when moving left, the car moves from t5 to t4 (distance = t4-t5, which is larger), t4 to t3 (distance = t3-t4, large), t3 to t2 (medium), t2 to t1 (small). Wait, that’s the opposite! So the distance covered (leftward) between t5-t4 is larger, t4-t3 is large, t3-t2 medium, t2-t1 small. So the speed (magnitude) is decreasing as it moves left (because distance covered per time interval decreases). Wait, this is the key mistake!

Oh no, I had the time direction wrong. Time increases from left to right in the diagram. So the first car (leftmost) is at time t1, the next at t2 (later time), then t3, t4, t5 (latest time). So the car moves from t1 (left) to t5 (right) over time. So the spacing between t1-t2 is small (short time, short distance), t2-t3 is medium, t3-t4 large, t4-t5 larger. So speed increases (speeding up right) because distance per time (speed) increases.

Now, if the car moves leftward (opposite direction), it’s moving from t5 (rightmost, latest time) to t4 to t3 to t2 to t1 (leftmost, earliest time) over time. So at t5 (position 5), then t4 (position 4, left of 5), t3 (position 3, left of 4), etc. The distance between t5 and t4 is the spacing between position 5 and 4 (which is larger, from original diagram: t4-t5 is larger), then t4-t3 (large), t3-t2 (medium), t2-t1 (small). So the distance covered (leftward) between t5-t4 is larger, t4-t3 is large, t3-t2 medium, t2-t1 small. Wait, that means the speed (magnitude) is decreasing as time increases (because distance per time interval decreases). So velocity is leftward (negative), and speed is decreasing (so velocity is becoming less negative, i.e., approaching zero). Thus, the change in velocity (Δv) is positive (since velocity goes from -large to -medium to -small, so Δv = v2 - v1 is positive). Acceleration is Δv/Δt, so positive (rightward).

Ah! That’s the correct approach. Let’s formalize:

- Let right be positive.
- Original motion (rightward): velocity is positive, and increasing (speeding up right), so acceleration is positive (right).
- Reversed motion (leftward): velocity is negative (left), and the speed (magnitude) is decreasing (because distance per time interval decreases as time increases). So velocity goes from -v5 to -v4 to -v3 to -v2 to -v1, where |v5| > |v4| > |v3| > |v2| > |v1|. Thus, velocity is becoming less negative (increasing toward zero), so Δv = v_new - v_old is positive (e.g., -v4 - (-v5) = v5 - v4 > 0).
- Acceleration = Δv/Δt, so positive (rightward).

Thus, if the car moves leftward in diagram D, it is slowing down (since speed decreases), so acceleration is opposite to velocity (velocity left, acceleration right).

# Brief Explanations:
When moving leftward, the car’s speed (magnitude of velocity) decreases (distance between positions decreases over time). Since velocity is left (negative) and speed decreases, velocity becomes less negative (approaches zero), so acceleration (Δv/Δt) is positive (rightward).
# Answer:
Direction of acceleration: Right
Explanation: The car moves leftward but slows down (speed decreases). Velocity (left) and acceleration (right) are opposite, so it decelerates (slows down) while moving left, meaning acceleration is right.

### Final Answers for Part 1 (Table Summary)

| Diagram | Reason (Brief)                     | Acceleration (Y/N) | Direction of Acceleration |
|---------|------------------------------------|--------------------|---------------------------|
| a       | Equal spacing (constant velocity) | N                  | (Blank)                   |
| b       | Spacing increases (speeding up R) | Y                  | Right                     |
| c       | Equal spacing (constant velocity) | N                  | (Blank)                   |
| d       | Spacing increases (speeding up R) | Y                  | Right                     |
| e       | Spacing decreases (slowing down R, so acceleration L) | Y | Left |

### Part 2 Answer
Direction of acceleration: Right
Explanation: When moving leftward, the car’s speed (magnitude of velocity) decreases (distance between positions decreases over time). Since velocity is left (negative) and speed decreases, velocity becomes less negative (approaches zero), so acceleration (Δv/Δt) is positive (rightward).

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QUESTION IMAGE

Question

Explanation:

Part 1: Analyzing Each Motion Diagram

Diagram a

Answer:

Diagram b