Speed Vs Time Graphs - Free Printable

Let's solve the problem step by step using the velocity-time graph provided for a London tube train journey between two stations.

---

1. Match the journey phase with its line on the graph

We are given three phases and need to match them with the correct segments of the graph.

#### The graph has three distinct parts:
- 0 to 30 seconds: Speed increases from 0 to 60 m/s → Acceleration
- 30 to 80 seconds: Speed remains constant at 60 m/s → Constant speed
- 80 to 140 seconds: Speed decreases from 60 m/s to 0 → Deceleration (retardation)

Now match the descriptions:

| Description | Graph Segment |
|-----------|---------------|
| Departs the station and increases speed for 30 seconds | 0–30 s (upward sloping line) |
| Travels at a constant speed of 60 m/s | 30–80 s (horizontal line) |
| Decreases speed to approach the next station | 80–140 s (downward sloping line) |

✔ So:
- "Departs..." → 0–30 s
- "Travels at constant speed..." → 30–80 s
- "Decreases speed..." → 80–140 s

---

2. Find the acceleration over the first 30 seconds

Acceleration is the slope of the velocity-time graph.

$$
\text{Acceleration} = \frac{\Delta v}{\Delta t} = \frac{v_f - v_i}{t_f - t_i}
$$

From 0 to 30 seconds:
- $ v_i = 0 $ m/s
- $ v_f = 60 $ m/s
- $ \Delta t = 30 $ s

$$
a = \frac{60 - 0}{30} = \frac{60}{30} = 2 \, \text{m/s}^2
$$

✔ Acceleration = 2 m/s²

---

3. Find the retardation over the final 70 seconds

Retardation is negative acceleration, or deceleration.

Final 70 seconds: from 80 s to 140 s (since 140 – 80 = 60 seconds). Wait — that’s only 60 seconds, not 70.

But the question says "final 70 seconds". Let's check:

Total time = 140 s
So final 70 seconds = from 70 s to 140 s?

But the speed is constant from 30–80 s, then starts decreasing at 80 s.

Wait — actually, the deceleration phase is from 80 s to 140 s, which is 60 seconds, not 70.

But the question asks for retardation over the final 70 seconds.

Let’s clarify: if the journey ends at 140 s, then the last 70 seconds would be from 70 s to 140 s.

So we must consider two parts:
- From 70 s to 80 s: constant speed (60 m/s)
- From 80 s to 140 s: deceleration

But retardation refers to deceleration, so we need to find the average retardation only during the deceleration phase? Or over the full 70 seconds?

However, since retardation is defined as rate of decrease in speed, it only applies when speed is decreasing.

So likely, the question intends the deceleration phase, but says "final 70 seconds" — perhaps a typo?

Let’s double-check the graph:

- Acceleration: 0–30 s
- Constant speed: 30–80 s → 50 seconds
- Deceleration: 80–140 s → 60 seconds

So the final 60 seconds are deceleration.

But the question says "final 70 seconds" — this may be an error, or perhaps they mean the last 60 seconds.

Alternatively, maybe they want the average retardation over the last 70 seconds, including some constant speed portion?

Let’s suppose it's a mistake and they mean the deceleration phase (80–140 s), which is 60 seconds.

But let's calculate both possibilities.

---

#### Option 1: Retardation over 80–140 s (60 seconds)

Speed changes from 60 m/s to 0 m/s in 60 seconds.

$$
\text{Retardation} = \frac{\Delta v}{\Delta t} = \frac{0 - 60}{140 - 80} = \frac{-60}{60} = -1 \, \text{m/s}^2
$$

So magnitude of retardation = 1 m/s²

✔ This is standard.

But the question says "final 70 seconds" — from 70 s to 140 s?

Let’s try that.

---

#### Option 2: Final 70 seconds = 70 s to 140 s

Break into two parts:
- 70 s to 80 s: 10 seconds at constant speed (no change in velocity)
- 80 s to 140 s: 60 seconds, speed drops from 60 to 0

So total change in velocity over 70 seconds = –60 m/s (since it only decreases after 80 s)

$$
\text{Average acceleration} = \frac{\Delta v}{\Delta t} = \frac{0 - 60}{70} = \frac{-60}{70} = -0.857 \, \text{m/s}^2
$$

But retardation typically refers to the magnitude of deceleration during the braking phase, not average over a period including constant speed.

Also, the graph shows deceleration starting at 80 s, so retardation should be calculated only during the actual slowing down phase.

Therefore, the most reasonable interpretation is that the "final 70 seconds" is a typo, and it should be final 60 seconds.

Alternatively, perhaps the graph is labeled differently?

Wait — let’s check the x-axis: Time (s) goes from 0 to 140, with ticks every 20 s.

- 0–30: acceleration
- 30–80: constant speed → 50 seconds
- 80–140: deceleration → 60 seconds

So yes, 60 seconds of deceleration.

Thus, likely the question meant "final 60 seconds", but wrote 70.

But let’s suppose it's correct as written: final 70 seconds → 70 s to 140 s.

Then:
- At 70 s: still in constant speed phase → speed = 60 m/s
- At 140 s: speed = 0 m/s
- So Δv = –60 m/s over 70 s

$$
a = \frac{-60}{70} = -0.857 \, \text{m/s}^2
\Rightarrow \text{Retardation} = 0.857 \, \text{m/s}^2
$$

But this includes a segment where there is no retardation (70–80 s), so the actual retardation (i.e., rate of slowing) is only during 80–140 s.

Hence, the correct interpretation is that retardation refers to the deceleration phase, i.e., 80–140 s, even though the question says “final 70 seconds”.

Possibility: maybe the graph ends at 130 s? But it clearly says 140 s.

Wait — look at the graph: the line goes from 80 to 140 s.

But let’s count the time:

- 80 to 140 = 60 seconds

So unless the graph is mislabeled, it's 60 seconds.

Given that, and since the question says "final 70 seconds", it might be a typo.

But to be safe, let’s assume the question means during the deceleration phase, and "70" is a typo.

Alternatively, perhaps the deceleration starts earlier?

No — from graph: flat line until 80 s, then slope down.

So deceleration begins at 80 s.

Thus, final 60 seconds.

But the question says 70 seconds — could it be that the total journey is longer?

No — graph ends at 140 s.

So either:
- It's a typo, and it should be 60 seconds, or
- We compute average retardation over the last 70 seconds, including constant speed.

But retardation is a physical quantity meaning deceleration, so it should be computed only during the slowing down phase.

Therefore, I will proceed with:

> Retardation = 1 m/s², assuming the intended time is 60 seconds (80–140 s).

But to answer exactly as asked:

---

#### If "final 70 seconds" means from 70 s to 140 s:

- Initial speed at 70 s: 60 m/s
- Final speed at 140 s: 0 m/s
- Change in velocity: –60 m/s
- Time interval: 70 s

$$
\text{Average acceleration} = \frac{-60}{70} = -0.857 \, \text{m/s}^2
\Rightarrow \text{Retardation} = 0.857 \, \text{m/s}^2
$$

But this is not the actual retardation — it's an average over a period that includes no deceleration.

So this is incorrect.

✔ Therefore, the question likely meant "final 60 seconds".

So we’ll go with:

> Retardation = 1 m/s²

(Answer based on 80–140 s: Δv = –60 m/s, Δt = 60 s)

---

4. Find the total distance travelled

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

We break the graph into three parts:

1. 0–30 s: Triangle
2. 30–80 s: Rectangle
3. 80–140 s: Triangle

---

#### Part 1: 0–30 s → Triangle

Base = 30 s, Height = 60 m/s

$$
\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 30 \times 60 = 900 \, \text{m}
$$

---

#### Part 2: 30–80 s → Rectangle

Width = 80 – 30 = 50 s
Height = 60 m/s

$$
\text{Area} = 50 \times 60 = 3000 \, \text{m}
$$

---

#### Part 3: 80–140 s → Triangle

Base = 140 – 80 = 60 s
Height = 60 m/s

$$
\text{Area} = \frac{1}{2} \times 60 \times 60 = 1800 \, \text{m}
$$

---

#### Total Distance:

$$
900 + 3000 + 1800 = 5700 \, \text{m}
$$

✔ Total distance = 5700 meters (or 5.7 km)

---

✔ Final Answers:

1. Matching:
- "Departs the station and increases speed for 30 seconds" → 0–30 s
- "Travels at constant speed of 60 m/s" → 30–80 s
- "Decreases speed to approach the next station" → 80–140 s

2. Acceleration over first 30 seconds:
$$
a = \frac{60 - 0}{30} = 2 \, \text{m/s}^2
$$

3. Retardation over final 70 seconds:
Since deceleration occurs from 80 to 140 s (60 seconds), and the question likely meant 60 seconds, we compute:
$$
\text{Retardation} = \frac{60 - 0}{60} = 1 \, \text{m/s}^2
$$
*(Note: If truly over 70 seconds, it would be ~0.857 m/s², but that includes non-decelerating phase — unlikely intended.)*

4. Total distance travelled:
$$
\text{Area} = 900 + 3000 + 1800 = \boxed{5700} \, \text{m}
$$

---

✔ Summary:

| Question | Answer |
|--------|--------|
| 1. Matching | As above |
| 2. Acceleration | $ 2 \, \text{m/s}^2 $ |
| 3. Retardation | $ 1 \, \text{m/s}^2 $ (assuming 60 s deceleration) |
| 4. Distance | $ 5700 \, \text{m} $ |

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