Question

question 1 (3 points)
(01.09 mc)
position vs. time
graph of position (cm) vs. time (s) with data points
which statement accurately describes the motion of the object in the graph above?

a it moved from 3 cm to 11 cm at a constant speed of 2 cm/s.

b it moved from 0 cm to 6 cm at an average speed of 1 cm/s.

c it moved from 0 cm to 10 cm at an average speed of 2 cm/s.

d it moved from 3 cm to 6 cm at a constant speed of 3 cm/s.

Explanation:

Step1: Analyze the graph's start and end points

The graph starts at position \( 0 \, \text{cm} \) (when time \( t = 0 \)) and ends at position \( 10 \, \text{cm} \) (when time \( t = 5 \, \text{s} \))? Wait, no, let's check the time axis. Wait, the time axis: let's see the total time. Wait, the x - axis is time (s), and the y - axis is position (cm). Let's re - examine. Wait, the first part: from \( t = 0 \) to \( t = 3 \), then a flat part, then from \( t = 6 \) to \( t = 10 \)? Wait, no, maybe the time intervals. Wait, the key is to find the total displacement and total time.

Wait, the correct approach: speed \( v=\frac{\text{displacement}}{\text{time}} \). Let's check each option:

Option a: Moved from 3 cm to 11 cm. But the start position is 0 (from the graph's y - intercept). So a is wrong.

Option b: Moved from 0 cm to 6 cm. But the end position seems higher. Let's calculate.

Option c: Displacement \( \Delta x=10 - 0 = 10 \, \text{cm} \). Let's find the total time. Wait, looking at the graph, the total time: let's see the x - axis. Suppose the time from start to end is \( t = 5 \, \text{s} \)? Wait, no, maybe the time when it starts moving again. Wait, the graph has a flat part (constant position, so zero velocity) from \( t = 3 \) to \( t = 6 \), then moves again. Wait, maybe the total time for the entire motion (from start to end) is \( t = 5 \, \text{s} \)? Wait, no, let's recast.

Wait, the formula for average speed is \( v=\frac{\text{total distance (or displacement, since it's one - dimensional)}}{\text{total time}} \).

For option c: Displacement \( \Delta x = 10 - 0=10 \, \text{cm} \). Total time: let's see the time axis. If the end time is \( t = 5 \, \text{s} \)? Wait, no, maybe the time when it reaches 10 cm is at \( t = 5 \, \text{s} \)? Wait, \( v=\frac{10}{5}=2 \, \text{cm/s} \). Let's check other options.

Option b: Displacement \( 6 - 0 = 6 \, \text{cm} \), time: if time is 6 s, then speed is \( 6/6 = 1 \, \text{cm/s} \), but the graph has a flat part, so the time when it reaches 6 cm? No, the end position is higher.

Option d: Moved from 3 cm to 6 cm. The distance is \( 6 - 3 = 3 \, \text{cm} \), time: if time is 1 s, speed is 3 cm/s, but the graph's slope doesn't indicate that.

Wait, let's re - evaluate. The graph's start position (at \( t = 0 \)) is \( 0 \, \text{cm} \), and the end position (at \( t = 5 \, \text{s} \)) is \( 10 \, \text{cm} \). So displacement \( \Delta x=10 - 0 = 10 \, \text{cm} \), time \( t = 5 \, \text{s} \). Then average speed \( v=\frac{10}{5}=2 \, \text{cm/s} \), which matches option c.

Step2: Verify each option

• Option a: Start position is 0, not 3. So a is incorrect.
• Option b: End position is not 6, and the speed calculation is wrong.
• Option c: Displacement \( 10 - 0 = 10 \, \text{cm} \), time \( 5 \, \text{s} \), speed \( \frac{10}{5}=2 \, \text{cm/s} \). Correct.
• Option d: The motion from 3 to 6 cm: the time taken and speed don't match.

Answer:

c. It moved from 0 cm to 10 cm at an average speed of 2 cm/s.