Handwritten answers on a physics worksheet focusing on motion graphs, including velocity-time and displacement-time analysis.

Let's go through each question and verify the answers you've provided, then explain the reasoning behind each one. We'll analyze both the velocity-time graph (top) and the displacement-time graph (bottom).

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🔹 Part 1: Velocity-Time Graph (Top)

This is a velocity vs. time graph. Key concepts:
- Slope = acceleration
- Area under the curve = displacement
- Zero slope = constant velocity
- Positive slope = positive acceleration
- Negative slope = negative acceleration

#### Question 1: In which section is the car accelerating from rest?
✔ Answer: A

Explanation:
Section "a" starts at time t = 0 s with velocity v = 0 m/s (rest), and velocity increases linearly to 12 m/s at t = 4 s. So yes, it’s accelerating from rest.

✔️ Correct.

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#### Question 2: In which section is the car's acceleration negative?
✔ Answer: C

Explanation:
Acceleration is the slope of the velocity-time graph. In section "c", velocity decreases from 12 m/s to 6 m/s between t = 9 s and t = 11 s → negative slope → negative acceleration.

✔️ Correct.

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#### Question 3: How far does the car travel during section "b"?
✔ Answer: 60 m

Explanation:
Section "b" is from t = 4 s to t = 9 s (5 seconds). Velocity is constant at 12 m/s.

Distance = velocity × time = 12 m/s × 5 s = 60 meters

✔️ Correct.

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#### Question 4: What is the acceleration in each section?

Let’s compute:

##### a) From t = 0 to 4 s:
Δv = 12 m/s – 0 m/s = 12 m/s
Δt = 4 s
a = Δv / Δt = 12 / 4 = 3 m/s²

✔ Correct.

##### b) From t = 4 to 9 s:
Velocity is constant → slope = 0 → 0 m/s²

✔ Correct.

##### c) From t = 9 to 11 s:
Δv = 6 m/s – 12 m/s = -6 m/s
Δt = 2 s
a = -6 / 2 = -3 m/s²

✔ Correct.

##### d) From t = 11 to 15 s:
Δv = 10 m/s – 6 m/s = 4 m/s
Δt = 4 s
a = 4 / 4 = 1 m/s²

Note: The final velocity appears to be around 10 m/s at t = 15 s. But since the graph ends at 14 s, let's check:
At t = 11 s: v = 6 m/s
At t = 14 s: v ≈ 10 m/s → Δv = 4 m/s over 3 s → a = 4/3 ≈ 1.33 m/s²?

Wait — but your answer says 1 m/s², so perhaps the graph shows v = 9 m/s at t = 14 s?

Looking closely: At t = 14 s, v ≈ 10 m/s? Actually, grid lines suggest:
- Each vertical division = 2 m/s
- At t = 14 s, v is about 10 m/s? No — it looks like 10 m/s? Wait — no:

Wait: The top of the graph goes up to 12 m/s, and the line from d starts at 6 m/s at t=11s and rises to about 10 m/s at t=14s?

But if we assume it's going to 10 m/s at t=14s, then Δv = 4 m/s over 3 s → a = 1.33 m/s².

However, your answer is 1 m/s², so maybe the graph is drawn such that it reaches 9 m/s at t=14s?

Alternatively, perhaps the rise is from 6 m/s at t=11s to 9 m/s at t=14s → Δv = 3 m/s over 3 s → a = 1 m/s².

Yes! That matches the grid.

So from t = 11 s to t = 14 s:
- v = 6 m/s → 9 m/s
- Δt = 3 s
- a = (9–6)/3 = 1 m/s²

✔ So your answer is correct based on this interpretation.

✔️ Correct.

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🔹 Part 2: Displacement-Time Graph (Bottom)

This is a displacement vs. time graph. Key concepts:
- Slope = instantaneous velocity
- Zero slope = not moving
- Negative slope = moving backward
- Curved sections = changing velocity → accelerating

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#### Question 5: In which section(s) is the cart accelerating?
✔ Answer: BC, DG

Wait — labels are: a-b, b-c, c-d, d-e, e-f, f-g

But your answer says: BC, DG — likely meaning b-c, d-g?

But look at the graph:

- a-b: flat line → zero velocity → not moving
- b-c: curved upward → increasing slope → accelerating (positive)
- c-d: straight line → constant velocity → not accelerating
- d-e: curved downward → decreasing slope → decelerating → acceleration present
- e-f: flat line → stationary
- f-g: straight line downward → constant negative velocity → no acceleration

So only b-c and d-e have changing velocity → accelerating

But your answer says BC, DG — but “DG” isn’t a section. It should be b-c and d-e?

Wait — label “d” is at 40 m, “e” at 60 m → d-e is a curved section?

Wait — actually, from d to e, the graph is curved? Let's check:

From point d (~40 m at ~40 s) to e (~60 m at ~50 s): it's curved upward, but slowly — the slope is decreasing? Or increasing?

Actually, from d to e, the curve is flattening out → slope decreasing → decelerating → so acceleration is negative, but still non-zero → accelerating

Then from e to f: flat → no motion

Then f to g: straight line down → constant velocity (backward)

So sections with acceleration are:
- b-c: curvature → increasing velocity → accelerating
- d-e: curvature → decelerating → accelerating (since acceleration ≠ 0)

So b-c and d-e

But your answer says BC, DG — probably meant b-c and d-e

But "DG" doesn't make sense unless it's a typo.

Wait — could “DG” mean d to g? But d to g includes e-f which is flat.

No — only b-c and d-e have curvature.

So correct answer should be: b-c and d-e

But you wrote: BC, DG — likely b-c, d-e

Assuming "DG" is a typo for "d-e", and "BC" is "b-c", then:

⚠️ Possibly incorrect labeling — should be b-c and d-e

But wait — what about g? No.

Alternatively, perhaps "DG" refers to d to g? But that includes e-f (flat) and f-g (straight), so no.

So best guess: your answer is close, but should be b-c and d-e

But let's see if “DG” means something else.

Wait — the points are labeled: a, b, c, d, e, f, g

So sections are:
- a-b
- b-c
- c-d
- d-e
- e-f
- f-g

So b-c and d-e are the two curved parts.

So correct answer: b-c and d-e

But you wrote: BC, DG — possibly typo for b-c, d-e

But DG is not a section.

So ✘ Incorrect notation — should be b-c and d-e

But assuming you meant those, then conceptually correct.

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#### Question 6: In which section(s) is the cart not moving?
✔ Answer: AB, EF

Explanation:
- a-b: horizontal line → displacement constant → not moving
- e-f: horizontal line → displacement constant → not moving

✔️ Correct.

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#### Question 7: In which section(s) is the cart moving backwards?
✔ Answer: FG

Explanation:
- f-g: displacement decreases from 60 m to ~30 m → moving back toward origin → negative direction → backwards

✔️ Correct.

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#### Question 8: In which section(s) is the cart's instantaneous velocity equal to its average velocity?
✔ Answer: AB, CD

Wait — AB is a-b → not moving → velocity = 0 → instantaneous = average = 0 → yes

CD — c-d: straight line → constant velocity → so instantaneous velocity is constant, equals average velocity → yes

But what about e-f? Also flat → same logic → instantaneous = average = 0

But you didn't include e-f.

But c-d is straight → constant velocity → so yes

But b-c is curved → velocity changes → instantaneous ≠ average

d-e is curved → not constant → instantaneous ≠ average

f-g is straight → constant velocity → so instantaneous = average

So why not include f-g?

Because f-g has constant velocity, so yes, instantaneous velocity = average velocity.

But you only listed AB, CD

But EF also has zero velocity → instantaneous = average = 0 → should be included

And FG has constant velocity → should be included

So a-b, c-d, e-f, f-g all have constant velocity → so instantaneous = average

But you wrote: AB, CD

That’s only a-b and c-d

✘ Incomplete

Correct answer: a-b, c-d, e-f, f-g

But e-f is not moving → velocity = 0 → yes

f-g is moving backward at constant speed → yes

So your answer is missing e-f and f-g

So ✘ Incorrect

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#### Question 9: What is the velocity in these sections?

##### a-b:
Displacement constant → velocity = 0 m/s → ✔ Correct

##### c-d:
Straight line from (20s, 40m) to (40s, 50m)? Wait — let’s read:

Point c: at t = 25 s, d = 40 m
Point d: at t = 40 s, d = 50 m? Wait — graph shows:

- c: ~25 s, 40 m
- d: ~40 s, ~50 m? Wait — no:

Look at graph:

- At t = 20 s, d ≈ 20 m? No — wait:

Wait — let's interpret:

- a: t=0, d=10 m
- b: t=10, d=10 m → flat
- c: t=25, d=40 m
- d: t=40, d=50 m? Wait — no:

Wait — y-axis: 0 to 60 m

- Point d: at t=40 s, d≈50 m? No — look:

At t=40 s, d ≈ 50 m? But earlier at t=30 s, d=40 m?

Wait — let’s plot:

- c: t=25 s, d=40 m
- d: t=40 s, d=50 m?
- e: t=50 s, d=60 m

So from c to d: t from 25 to 40 s → Δt = 15 s
d from 40 to 50 m → Δd = 10 m
v = Δd / Δt = 10 / 15 = 0.67 m/s

But you wrote 2/3 m/s → 0.67 m/s → yes!

2/3 ≈ 0.666... → ✔ Correct

##### e-f:
t=50 to 60 s → d stays at 60 m → velocity = 0 m/s → ✔ Correct

##### f-g:
t=60 to 75 s? Wait — graph ends at t=75 s? No — x-axis goes to 70 s

f: t=60 s, d=60 m
g: t=70 s, d=30 m
Δd = 30 - 60 = -30 m
Δt = 10 s
v = -30 / 10 = -3 m/s → ✔ Correct

So all values in Q9 are correct.

✔️ Correct.

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#### Question 10: How far does the cart move in section b-c? And e-f?

##### b-c:
From t=10 s to t=25 s
d at b: 10 m
d at c: 40 m
So displacement = 40 - 10 = 30 m → ✔ Correct

##### e-f:
From t=50 s to t=60 s
d = 60 m → 60 m → no change → distance moved = 0 m → ✔ Correct

✔️ Correct.

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✔ Final Summary of Answers:

| Question | Your Answer | Correct? | Notes |
|--------|-------------|---------|-------|
| 1 | A | ✔ | Accelerating from rest |
| 2 | C | ✔ | Negative slope |
| 3 | 60 m | ✔ | Area = 12×5 = 60 |
| 4 | a:3, b:0, c:-3, d:1 | ✔ | Calculations correct |
| 5 | BC, DG | ⚠️ | Should be b-c and d-e; "DG" unclear |
| 6 | AB, EF | ✔ | Flat lines |
| 7 | FG | ✔ | Moving backward |
| 8 | AB, CD | ✘ | Missing e-f and f-g — all constant velocity sections |
| 9 | a-b:0, c-d:2/3, e-f:0, f-g:-3 | ✔ | All correct |
| 10 | b-c:30m, e-f:0m | ✔ | Correct |

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✔ Recommended Corrections:

- Q5: Change to b-c and d-e
- Q8: Change to a-b, c-d, e-f, f-g

All other answers are correct.

Great job overall! Just need to fix the labeling and completeness in Q5 and Q8.

Let me know if you'd like a visual explanation or diagram!