Position-time and velocity-time graphs review worksheet with questions and blank graph for student analysis.

Let’s solve each question step by step using the position-time graph.

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Question 1: Describe the motion of the object from...

We look at how the position changes over time on the graph.

a. 0–2 seconds:
The line is flat (horizontal) at position = 5 m. That means the object is not moving — it’s staying still.
→ *At rest*

b. 2–5 seconds:
The line goes up from 5 m to 20 m in 3 seconds. It’s a straight diagonal line going up → constant speed forward.
→ *Moving forward at constant speed*

c. 5–8 seconds:
Line is flat again at 20 m → not moving.
→ *At rest*

d. 8–9 seconds:
Line goes down from 20 m to 15 m in 1 second → moving backward at constant speed.
→ *Moving backward at constant speed*

e. 9–12 seconds:
Line goes down from 15 m to 0 m in 3 seconds → still moving backward, but slower than before? Wait — let’s check slope.

From 9 to 12 sec: Δposition = 0 - 15 = -15 m, Δtime = 3 s → velocity = -5 m/s
From 8 to 9 sec: Δposition = 15 - 20 = -5 m, Δtime = 1 s → velocity = -5 m/s
Same speed! So actually, from 8 to 12 seconds, it’s moving backward at same constant speed. But the question breaks it into d and e.

So for e: still moving backward at constant speed until it reaches origin.

But since the graph shows two segments (8-9 and 9-12), we describe them separately as asked.

Actually, looking again: from 8 to 9: drops 5m in 1s → steep drop
from 9 to 12: drops 15m in 3s → same steepness? Let’s calculate slope:

Segment 8-9: (15 - 20)/(9 - 8) = -5/1 = -5 m/s
Segment 9-12: (0 - 15)/(12 - 9) = -15/3 = -5 m/s → same velocity!

So actually, from 8 to 12 seconds, it’s one continuous motion backward at -5 m/s. But the graph has a point at t=9, maybe just to mark? The line is straight from 8 to 12? Wait — no, looking at the graph description: it says “from 8 to 9” then “9 to 12”, and the line is broken at 9? Actually, in the original problem, the graph likely has a kink at t=9? But according to calculation, if both segments have same slope, it’s straight.

Wait — let me recheck with actual values from typical such graphs. Usually in these problems, from 8 to 9 it might be steeper? But here, based on coordinates:

Assume from graph:
t=8, x=20
t=9, x=15
t=12, x=0

So yes, from 8 to 9: delta x = -5, delta t = 1 → v = -5 m/s
from 9 to 12: delta x = -15, delta t = 3 → v = -5 m/s → same.

So perhaps the graph is drawn with a point at t=9 but it's still a straight line? Or maybe it's two segments with same slope. Either way, motion is same.

But since the question asks to describe each interval, we’ll say:

d. 8–9 seconds: Moving backward at constant speed
e. 9–12 seconds: Still moving backward at same constant speed until reaching start

But to be precise, let’s write:

a. 0–2 s: At rest (position constant at 5 m)
b. 2–5 s: Moving forward at constant speed (from 5 m to 20 m)
c. 5–8 s: At rest (position constant at 20 m)
d. 8–9 s: Moving backward at constant speed (from 20 m to 15 m)
e. 9–12 s: Continuing to move backward at same constant speed (from 15 m to 0 m)

Note: Since speed is same in d and e, you could say "moving backward at constant speed" for both, but we follow the intervals given.

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Question 2: What is the total distance traveled?

Distance is total path length, regardless of direction.

Break into segments:

0–2 s: no movement → 0 m
2–5 s: from 5 m to 20 m → |20 - 5| = 15 m
5–8 s: no movement → 0 m
8–9 s: from 20 m to 15 m → |15 - 20| = 5 m
9–12 s: from 15 m to 0 m → |0 - 15| = 15 m

Total distance = 0 + 15 + 0 + 5 + 15 = 35 meters

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Question 3: What is the displacement of the object?

Displacement = final position - initial position

Initial position at t=0: 5 m
Final position at t=12: 0 m
Displacement = 0 - 5 = -5 meters

(Or 5 meters in the negative direction)

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Question 4: What is the average speed of the object?

Average speed = total distance / total time

Total distance = 35 m (from Q2)
Total time = 12 seconds

Average speed = 35 / 12 ≈ 2.92 m/s (we can leave as fraction or decimal)

35 ÷ 12 = 2.9166... → round to 2.92 m/s or keep as 35/12 m/s

Usually in school, they accept decimal rounded to two places.

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Question 5: What is the average velocity of the object?

Average velocity = displacement / total time

Displacement = -5 m (from Q3)
Total time = 12 s

Average velocity = -5 / 12 ≈ -0.42 m/s

(Again, or -5/12 m/s)

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Question 6: Create a velocity-time graph from the position-time graph.

We need to find velocity for each time interval. Velocity is slope of position-time graph.

Intervals:

0–2 s: slope = 0 → velocity = 0 m/s
2–5 s: slope = (20 - 5)/(5 - 2) = 15/3 = 5 m/s → velocity = +5 m/s
5–8 s: slope = 0 → velocity = 0 m/s
8–12 s: slope = (0 - 20)/(12 - 8) = (-20)/4 = -5 m/s → but wait, earlier we saw from 8-9 and 9-12 both -5 m/s, so overall from 8 to 12, velocity is constant -5 m/s.

Actually, since from 8 to 12 it's a straight line with slope -5 m/s, we can treat it as one segment.

But the position graph may have a point at t=9, but if the line is straight from 8 to 12, then velocity is constant -5 m/s throughout 8-12.

In the problem, question 1 separates 8-9 and 9-12, but for velocity graph, if the slope is same, it's one horizontal line.

To be safe, let's compute per interval as defined in Q1.

For velocity-time graph:

- From t=0 to t=2: v = 0 m/s
- From t=2 to t=5: v = +5 m/s
- From t=5 to t=8: v = 0 m/s
- From t=8 to t=9: v = (15-20)/(9-8) = -5 m/s
- From t=9 to t=12: v = (0-15)/(12-9) = -5 m/s → same as above

So from t=8 to t=12, velocity is constantly -5 m/s. We can draw a single horizontal line from t=8 to t=12 at v=-5 m/s.

So velocity-time graph will have:

- Horizontal line at v=0 from t=0 to 2
- Jump to v=5 from t=2 to 5
- Drop to v=0 from t=5 to 8
- Drop to v=-5 from t=8 to 12

All are horizontal lines (constant velocity in each interval).

Now, to sketch this on the provided grid (which we don't have, but we describe):

On velocity-time axes:

Time (x-axis) from 0 to 12 s
Velocity (y-axis) probably from -6 to 6 or similar.

Plot:

- [0,2]: v=0 → line on x-axis
- [2,5]: v=5 → horizontal line at y=5
- [5,8]: v=0 → back to x-axis
- [8,12]: v=-5 → horizontal line at y=-5

At the transition points (t=2,5,8), there are jumps (discontinuities), which is fine for piecewise constant velocity.

So the graph consists of four horizontal segments.

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Now, compiling all answers.

Final Answer:

1. Motion description:
a. 0–2 s: At rest
b. 2–5 s: Moving forward at constant speed
c. 5–8 s: At rest
d. 8–9 s: Moving backward at constant speed
e. 9–12 s: Moving backward at constant speed (same as d)

2. Total distance traveled: 35 meters

3. Displacement: -5 meters

4. Average speed: 35/12 m/s ≈ 2.92 m/s

5. Average velocity: -5/12 m/s ≈ -0.42 m/s

6. Velocity-time graph:
- From t=0 to 2 s: velocity = 0 m/s
- From t=2 to 5 s: velocity = +5 m/s
- From t=5 to 8 s: velocity = 0 m/s
- From t=8 to 12 s: velocity = -5 m/s
(Draw horizontal lines at these velocities for each interval)