Worksheet for calculating acceleration using the formula a = (vf - vi) / t.

Let's solve each problem step by step using the formula for acceleration:

$$
a = \frac{v_f - v_i}{t}
$$

Where:
- $ a $ = acceleration (in m/s² or km/hr/s, depending on units)
- $ v_f $ = final velocity
- $ v_i $ = initial velocity
- $ t $ = time

We’ll need to convert units where necessary (especially km/hr to m/s or seconds), and ensure all units are consistent.

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Problems 1–5: Calculate Acceleration

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#### 1.
- Initial velocity ($v_i$) = 0 km/hr
- Final velocity ($v_f$) = 24 km/hr
- Time ($t$) = 3 s

But we have km/hr and seconds, so convert km/hr to m/s first.

> Conversion: $ 1 \text{ km/hr} = \frac{1000}{3600} = \frac{5}{18} \text{ m/s} $

So:
- $ v_i = 0 $ km/hr = 0 m/s
- $ v_f = 24 \times \frac{5}{18} = \frac{120}{18} = 6.67 $ m/s
- $ t = 3 $ s

Now calculate:
$$
a = \frac{6.67 - 0}{3} = \frac{6.67}{3} = 2.22 \text{ m/s}^2
$$

✔ Answer: $ \boxed{2.22 \text{ m/s}^2} $

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#### 2.
- $ v_i = 0 $ m/s
- $ v_f = 35 $ m/s
- $ t = 5 $ s

$$
a = \frac{35 - 0}{5} = \frac{35}{5} = 7 \text{ m/s}^2
$$

✔ Answer: $ \boxed{7 \text{ m/s}^2} $

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#### 3.
- $ v_i = 20 $ km/hr → $ 20 \times \frac{5}{18} = 5.56 $ m/s
- $ v_f = 60 $ km/hr → $ 60 \times \frac{5}{18} = 16.67 $ m/s
- $ t = 10 $ s

$$
a = \frac{16.67 - 5.56}{10} = \frac{11.11}{10} = 1.11 \text{ m/s}^2
$$

✔ Answer: $ \boxed{1.11 \text{ m/s}^2} $

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#### 4.
- $ v_i = 50 $ m/s
- $ v_f = 150 $ m/s
- $ t = 5 $ s

$$
a = \frac{150 - 50}{5} = \frac{100}{5} = 20 \text{ m/s}^2
$$

✔ Answer: $ \boxed{20 \text{ m/s}^2} $

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#### 5.
- $ v_i = 25 $ km/hr → $ 25 \times \frac{5}{18} = 6.94 $ m/s
- $ v_f = 1200 $ km/hr → $ 1200 \times \frac{5}{18} = 333.33 $ m/s
- $ t = 2 $ min = $ 120 $ s

$$
a = \frac{333.33 - 6.94}{120} = \frac{326.39}{120} = 2.72 \text{ m/s}^2
$$

✔ Answer: $ \boxed{2.72 \text{ m/s}^2} $

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Problem 6:

A car accelerates from standstill (0 km/hr) to 60 km/hr in 10.0 seconds.

Convert 60 km/hr to m/s:
$$
60 \times \frac{5}{18} = 16.67 \text{ m/s}
$$

$$
a = \frac{16.67 - 0}{10} = 1.67 \text{ m/s}^2
$$

✔ Answer: $ \boxed{1.67 \text{ m/s}^2} $

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Problem 7:

From 25 km/hr to 55 km/hr in 30 seconds.

Convert velocities:
- $ v_i = 25 \times \frac{5}{18} = 6.94 $ m/s
- $ v_f = 55 \times \frac{5}{18} = 15.28 $ m/s
- $ t = 30 $ s

$$
a = \frac{15.28 - 6.94}{30} = \frac{8.34}{30} = 0.278 \text{ m/s}^2
$$

✔ Answer: $ \boxed{0.278 \text{ m/s}^2} $

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Problem 8:

Train accelerating at 2.0 km/hr/s, initial velocity = 20 km/hr, time = 30 s.

Acceleration is given in km/hr per second, so use the same units.

Use:
$$
v_f = v_i + a \cdot t
$$

$$
v_f = 20 + (2.0 \text{ km/hr/s}) \times 30 \text{ s} = 20 + 60 = 80 \text{ km/hr}
$$

✔ Answer: $ \boxed{80 \text{ km/hr}} $

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Problem 9:

Runner reaches 11.1 m/s in 9 seconds. Starts from rest? Not stated, but likely $ v_i = 0 $.

Assume $ v_i = 0 $ m/s, $ v_f = 11.1 $ m/s, $ t = 9 $ s

$$
a = \frac{11.1 - 0}{9} = 1.23 \text{ m/s}^2
$$

Now find distance covered. Use:
$$
d = v_i t + \frac{1}{2} a t^2
$$

Since $ v_i = 0 $:
$$
d = \frac{1}{2} \times 1.23 \times (9)^2 = 0.615 \times 81 = 49.815 \text{ m}
$$

Alternatively, average velocity method:
$$
\text{Average velocity} = \frac{0 + 11.1}{2} = 5.55 \text{ m/s}
$$
$$
d = 5.55 \times 9 = 49.95 \text{ m} \quad (\text{slight rounding difference})
$$

Using precise $ a = 1.233... $, $ d = 0.5 \times 1.233 \times 81 = 49.95 $ m

✔ Acceleration: $ \boxed{1.23 \text{ m/s}^2} $
✔ Distance: $ \boxed{49.95 \text{ m}} $ (or ~50 m)

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✔ Final Answers Summary:

| Problem | Answer |
|--------|--------|
| 1 | 2.22 m/s² |
| 2 | 7 m/s² |
| 3 | 1.11 m/s² |
| 4 | 20 m/s² |
| 5 | 2.72 m/s² |
| 6 | 1.67 m/s² |
| 7 | 0.278 m/s² |
| 8 | 80 km/hr |
| 9 | Acceleration: 1.23 m/s²; Distance: 49.95 m |

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🔍 Notes:

- Always check unit consistency: Convert km/hr → m/s using $ \times \frac{5}{18} $
- Time must be in seconds when working with m/s²
- For problems involving acceleration in km/hr/s, you can keep units in km/hr and seconds — just be careful about what the answer means.

Let me know if you'd like this formatted as a completed worksheet!