Acceleration Calculations Answer Key | PDF - Free Printable

It looks like you've uploaded a worksheet titled "CALCULATIONS" that focuses on calculating acceleration using the formula:

\[
a = \frac{v_f - v_i}{t}
\]

Where:
- \( a \) = acceleration
- \( v_f \) = final velocity
- \( v_i \) = initial velocity
- \( t \) = time

The worksheet includes several problems, some of which are already partially filled in. Let's go through each problem step by step and explain how to solve them correctly.

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Problem 1:

- Initial Velocity (\(v_i\)) = 0 m/s
- Final Velocity (\(v_f\)) = 38 m/s
- Time (\(t\)) = 19 s

\[
a = \frac{38 - 0}{19} = \frac{38}{19} = 2.0 \, \text{m/s}^2
\]

✔ Correct answer: \(2.0 \, \text{m/s}^2\)

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Problem 2:

- \(v_i = 40 \, \text{m/s}\), \(v_f = 60 \, \text{m/s}\), \(t = 10 \, \text{s}\)

\[
a = \frac{60 - 40}{10} = \frac{20}{10} = 2.0 \, \text{m/s}^2
\]

✔ Correct answer: \(2.0 \, \text{m/s}^2\)

---

Problem 3:

- \(v_i = 50 \, \text{m/s}\), \(v_f = 25 \, \text{m/s}\), \(t = 5 \, \text{s}\)

\[
a = \frac{25 - 50}{5} = \frac{-25}{5} = -5.0 \, \text{m/s}^2
\]

Negative sign indicates deceleration (slowing down).
✔ Answer: \(-5.0 \, \text{m/s}^2\)

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Problem 4:

- A car accelerates from 0 km/h to 40 km/h in 20 seconds.
We need to convert km/h to m/s.

Conversion:
\[
40 \, \text{km/h} = \frac{40 \times 1000}{3600} = \frac{40}{3.6} \approx 11.11 \, \text{m/s}
\]

Now calculate acceleration:
\[
a = \frac{11.11 - 0}{20} = \frac{11.11}{20} \approx 0.555 \, \text{m/s}^2
\]

But the student wrote: \( \frac{40 \, \text{km/h}}{20 \, \text{s}} = 2 \, \text{km/h/s} \)

That’s correct in units, but not standard SI units.

To express in m/s²:
\[
2 \, \text{km/h/s} = \frac{2 \times 1000}{3600} = \frac{2}{3.6} \approx 0.556 \, \text{m/s}^2
\]

So both answers are acceptable depending on units, but SI units (m/s²) are preferred.

✔ Answer: \(0.556 \, \text{m/s}^2\) or \(2 \, \text{km/h/s}\)

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Problem 5:

- Accelerates from 25 km/h to 55 km/h in 30 seconds.

Convert velocities:
\[
25 \, \text{km/h} = \frac{25}{3.6} \approx 6.94 \, \text{m/s}, \quad 55 \, \text{km/h} = \frac{55}{3.6} \approx 15.28 \, \text{m/s}
\]

\[
a = \frac{15.28 - 6.94}{30} = \frac{8.34}{30} \approx 0.278 \, \text{m/s}^2
\]

Alternatively, in km/h/s:
\[
a = \frac{55 - 25}{30} = \frac{30}{30} = 1 \, \text{km/h/s}
\]

✔ Answer: \(1 \, \text{km/h/s}\) or \(0.278 \, \text{m/s}^2\)

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Problem 6:

- Train accelerates at \(2.0 \, \text{m/s}^2\) from rest (\(v_i = 0\)) for 10 seconds.
- Find final velocity: \(v_f = v_i + at = 0 + (2.0)(10) = 20 \, \text{m/s}\)

✔ Answer: \(20 \, \text{m/s}\)

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Problem 7:

- Runner reaches \(v = 11 \, \text{m/s}\) in \(t = 4 \, \text{s}\), starting from rest.

\[
a = \frac{11 - 0}{4} = 2.75 \, \text{m/s}^2
\]

Now find distance covered using:
\[
d = v_i t + \frac{1}{2} a t^2 = 0 + \frac{1}{2}(2.75)(4)^2 = \frac{1}{2}(2.75)(16) = 22 \, \text{m}
\]

✔ Acceleration: \(2.75 \, \text{m/s}^2\)
✔ Distance: \(22 \, \text{m}\)

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Summary of Correct Answers:

| Problem | Answer |
|--------|--------|
| 1 | \(2.0 \, \text{m/s}^2\) |
| 2 | \(2.0 \, \text{m/s}^2\) |
| 3 | \(-5.0 \, \text{m/s}^2\) |
| 4 | \(0.556 \, \text{m/s}^2\) or \(2 \, \text{km/h/s}\) |
| 5 | \(0.278 \, \text{m/s}^2\) or \(1 \, \text{km/h/s}\) |
| 6 | \(20 \, \text{m/s}\) |
| 7 | Acceleration: \(2.75 \, \text{m/s}^2\), Distance: \(22 \, \text{m}\) |

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Key Concepts:

- Acceleration is the rate of change of velocity over time.
- Always ensure units match — convert km/h to m/s when needed.
- Use kinematic equations:
- \( a = \frac{\Delta v}{\Delta t} \)
- \( v_f = v_i + at \)
- \( d = v_i t + \frac{1}{2} a t^2 \)

Let me know if you'd like this formatted as a printable solution sheet!