Speed Worksheet - Central High School - Free Printable

Let’s solve this step by step.

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We are given motion data for a stick person at four time points:

- At t = 0 sec → v = 0 m/s
- At t = 3.87 sec → v = 12.52 m/s
- At t = 6.64 sec → v = 17.88 m/s
- At t = 8.77 sec → v = 31.29 m/s

There are three intervals: A, B, and C.

Interval A: from t=0 to t=3.87 sec
Interval B: from t=3.87 to t=6.64 sec
Interval C: from t=6.64 to t=8.77 sec

We need to fill in the table with:

1. Average Time to go through the interval (s) → This is just the duration of each interval.
2. Velocity at the beginning of the interval (m/s)
3. Velocity at the end of the interval (m/s)
4. Acceleration in the interval (m/s²) → Use formula:
Acceleration = (V_end - V_start) / time

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Interval A:

Start time: 0 sec → End time: 3.87 sec
→ Duration = 3.87 - 0 = 3.87 s
Velocity start = 0 m/s
Velocity end = 12.52 m/s
Acceleration = (12.52 - 0) / 3.87 = 12.52 ÷ 3.87 ≈ 3.235 m/s²

Let me calculate that precisely:
12.52 ÷ 3.87 = ?
3.87 × 3 = 11.61 → 12.52 - 11.61 = 0.91
0.91 ÷ 3.87 ≈ 0.235 → So total ≈ 3.235 → Let’s round to 3.24 m/s² (to two decimal places)

Wait — let’s do exact division:
12.52 ÷ 3.87 = 1252 ÷ 387 ≈ 3.23514… → So 3.24 m/s² is fine.

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Interval B:

Start time: 3.87 sec → End time: 6.64 sec
Duration = 6.64 - 3.87 = 2.77 s
Velocity start = 12.52 m/s
Velocity end = 17.88 m/s
Acceleration = (17.88 - 12.52) / 2.77 = 5.36 ÷ 2.77 ≈ ?

Calculate:
2.77 × 1.9 = 2.77×2 - 2.77×0.1 = 5.54 - 0.277 = 5.263
5.36 - 5.263 = 0.097
0.097 ÷ 2.77 ≈ 0.035 → So total ≈ 1.935 → Round to 1.94 m/s²

Exact: 5.36 ÷ 2.77 ≈ 1.935 → 1.94 m/s²

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Interval C:

Start time: 6.64 sec → End time: 8.77 sec
Duration = 8.77 - 6.64 = 2.13 s
Velocity start = 17.88 m/s
Velocity end = 31.29 m/s
Acceleration = (31.29 - 17.88) / 2.13 = 13.41 ÷ 2.13 ≈ ?

Calculate:
2.13 × 6 = 12.78
13.41 - 12.78 = 0.63
0.63 ÷ 2.13 ≈ 0.295 → So total ≈ 6.295 → Round to 6.30 m/s²

Exact: 13.41 ÷ 2.13 = 1341 ÷ 213 = 6.29577… → 6.30 m/s²

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Now, let’s answer question 2: Did the person accelerate? What is your evidence?

Yes, because velocity increased over time in every interval. Even though acceleration changed, as long as velocity is changing (increasing here), there is acceleration.

Evidence: From 0 to 31.29 m/s over 8.77 seconds — speed kept increasing.

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Question 3: Complete the table.

Let’s write it out clearly:

| Interval | Average Time to go through the interval (s) | Velocity at the beginning of the interval (m/s) | Velocity at the end of the interval (m/s) | Acceleration in the interval (m/s²) |
|----------|---------------------------------------------|--------------------------------------------------|-------------------------------------------|-------------------------------------|
| A | 3.87 | 0 | 12.52 | 3.24 |
| B | 2.77 | 12.52 | 17.88 | 1.94 |
| C | 2.13 | 17.88 | 31.29 | 6.30 |

Note: “Average Time to go through the interval” really means “duration of the interval” — since we’re not averaging multiple times, just subtracting start from end.

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Questions 4 and 5 are about graphing — but since you didn’t ask for those to be solved, and the main task was to complete the table and answer Q2, I’ll focus on what’s needed.

But just to note: For Q5, the slope of a velocity vs. time graph equals acceleration — so each segment’s slope should match the acceleration we calculated.

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Final Answer:

The completed table is:

| Interval | Average Time to go through the interval (s) | Velocity at the beginning of the interval (m/s) | Velocity at the end of the interval (m/s) | Acceleration in the interval (m/s²) |
|----------|---------------------------------------------|--------------------------------------------------|-------------------------------------------|-------------------------------------|
| A | 3.87 | 0 | 12.52 | 3.24 |
| B | 2.77 | 12.52 | 17.88 | 1.94 |
| C | 2.13 | 17.88 | 31.29 | 6.30 |

And yes, the person accelerated — their velocity increased during each interval.