Physics worksheet on motion graphs with distance-time and velocity-time graphs for analyzing motion.

Let's solve each question step by step with clear explanations.

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Question 1: Distance-Time Graph – Guza’s Shopping Trip

#### Graph Analysis
The graph shows distance (in meters) vs. time (in seconds). The slope of the graph gives speed (since speed = distance/time).

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#### a) Describe the first 260 seconds of Guza’s trip

We break it down into segments:

- OA (0–50 s):
- Distance increases from 0 to 40 m.
- Straight line, so constant speed.
→ Guza is moving at a constant speed away from the starting point.

- AB (50–80 s):
- Distance remains constant at 40 m.
- Horizontal line → no change in position.
→ Guza is stationary (stopped).

- BC (80–150 s):
- Distance increases from 40 m to 80 m.
- Straight line, positive slope → constant speed.
→ Guza is moving at a constant speed away from the start.

- CD (150–240 s):
- Distance stays at 80 m.
- Horizontal line → no movement.
→ Guza is stationary.

- DE (240–260 s):
- Distance drops from 80 m to 40 m.
- Steep negative slope → moving back toward the start.
→ Guza is moving at a constant speed towards the starting point.

> ✔ So, in the first 260 seconds:
> - OA: Moving away at constant speed
> - AB: Stopped
> - BC: Moving away at constant speed
> - CD: Stopped
> - DE: Moving back toward start at constant speed

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#### b) In which region was Guza travelling the fastest?

Speed = slope of the distance-time graph.

Let’s calculate slopes for the moving sections:

- OA:
Δd = 40 m, Δt = 50 s → speed = 40/50 = 0.8 m/s

- BC:
Δd = 80 − 40 = 40 m, Δt = 150 − 80 = 70 s → speed = 40/70 ≈ 0.57 m/s

- DE:
Δd = 40 m (from 80 to 40), Δt = 260 − 240 = 20 s → speed = 40/20 = 2.0 m/s

✔ So, DE has the highest speed → Guza was travelling fastest during DE.

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#### c) Calculate Guza’s speed at:

i. OA:
Speed = Δd / Δt = (40 − 0)/(50 − 0) = 40/50 = 0.8 m/s

ii. BC:
Speed = (80 − 40)/(150 − 80) = 40/70 ≈ 0.57 m/s

iii. DE:
Speed = |(40 − 80)/(260 − 240)| = |-40/20| = 2.0 m/s

> ⬇️ Final Answers:
> - i. OA: 0.8 m/s
> - ii. BC: 0.57 m/s
> - iii. DE: 2.0 m/s

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Question 2: Velocity-Time Graph – Train Motion

This graph shows velocity (m/s) vs. time (s).

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#### a) Describe the motion of the train

- OA (0–60 s):
Velocity increases from 0 to 20 m/s.
→ Accelerating uniformly (straight line)

- AB (60–180 s):
Velocity remains constant at 20 m/s.
→ Moving at constant velocity

- BC (180–240 s):
Velocity increases from 20 to 25 m/s.
→ Accelerating again

- CD (240–360 s):
Velocity remains constant at 25 m/s.
→ Constant velocity (fastest part)

- DE (360–600 s):
Velocity decreases from 25 to 0 m/s.
→ Decelerating uniformly

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#### b) Calculate the greatest acceleration

Acceleration = gradient of velocity-time graph.

- OA:
Δv = 20 m/s, Δt = 60 s → a = 20/60 ≈ 0.33 m/s²

- BC:
Δv = 25 − 20 = 5 m/s, Δt = 240 − 180 = 60 s → a = 5/60 ≈ 0.083 m/s²

- DE (deceleration):
Δv = 0 − 25 = −25 m/s, Δt = 600 − 360 = 240 s → a = −25/240 ≈ −0.104 m/s²

So, the greatest magnitude of acceleration is in OA → 0.33 m/s²

> ✔ Greatest acceleration: 0.33 m/s² (during OA)

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#### c) Calculate the deceleration of the train

Deceleration = magnitude of acceleration during DE.

From above:
- a = (0 − 25)/(600 − 360) = −25/240 = −0.104 m/s²
→ Deceleration = 0.104 m/s²

> ✔ Deceleration = 0.104 m/s²

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#### d) Calculate the total distance travelled

On a velocity-time graph, distance = area under the graph.

We split the area into shapes:

1. Triangle OA (0–60 s):
Area = ½ × base × height = ½ × 60 × 20 = 600 m

2. Rectangle AB (60–180 s):
Width = 120 s, Height = 20 m/s → Area = 120 × 20 = 2400 m

3. Triangle BC (180–240 s):
Base = 60 s, Height = 5 m/s → Area = ½ × 60 × 5 = 150 m

4. Rectangle CD (240–360 s):
Width = 120 s, Height = 25 m/s → Area = 120 × 25 = 3000 m

5. Triangle DE (360–600 s):
Base = 240 s, Height = 25 m/s → Area = ½ × 240 × 25 = 3000 m

Total distance = 600 + 2400 + 150 + 3000 + 3000 = 9150 m

> ✔ Total distance travelled = 9150 meters

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Question 3: Complete the Table

| Feature on graph | Distance-Time | Velocity-Time |
|------------------------|-----------------------------------|-------------------------------------|
| Gradient | Speed (or rate of change of distance) | Acceleration (rate of change of velocity) |
| Horizontal Line | Stationary (not moving) | Constant velocity (no acceleration) |
| Upward Line | Moving away at increasing speed | Accelerating |
| Downward Line | Moving back or slowing down | Decelerating (negative acceleration) |
| Area under the Graph | NOT APPLICABLE | Distance travelled |

> ✔ Explanation:
> - Distance-Time: Gradient = speed; horizontal = stopped.
> - Velocity-Time: Gradient = acceleration; area = distance.
> - Upward/downward lines show increasing/decreasing velocity.

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✔ Final Answers Summary:

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Q1:

a)
- OA: Moving at constant speed away
- AB: Stationary
- BC: Moving at constant speed away
- CD: Stationary
- DE: Moving back at constant speed

b) Fastest in DE

c)
i. OA: 0.8 m/s
ii. BC: 0.57 m/s
iii. DE: 2.0 m/s

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Q2:

a)
- OA: Accelerating
- AB: Constant velocity
- BC: Accelerating
- CD: Constant velocity
- DE: Decelerating

b) Greatest acceleration: 0.33 m/s² (OA)

c) Deceleration: 0.104 m/s²

d) Total distance: 9150 m

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Q3: Table Completed

| Feature on graph | Distance-Time | Velocity-Time |
|------------------------|-------------------------------------------|--------------------------------------------|
| Gradient | Speed | Acceleration |
| Horizontal Line | Stationary | Constant velocity |
| Upward Line | Moving away at increasing speed | Accelerating |
| Downward Line | Moving back or slowing down | Decelerating |
| Area under the Graph | NOT APPLICABLE | Distance travelled |

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