Worksheet on calculating acceleration with examples and formulas.

Looking at the image, you’ve uploaded a worksheet titled “CALCULATIONS” that focuses on calculating acceleration using the formula:

> a = (v₂ – v₁) / t

Where:
- a = acceleration
- v₂ = final velocity
- v₁ = initial velocity
- t = time

The worksheet includes several problems (6–9), and some have handwritten answers. I’ll solve each problem step-by-step and explain the reasoning.

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🔢 Problem 6:

> A car accelerates from a standstill to 40 km/hr in 10 seconds. What is its acceleration?

Given:
- Initial velocity (v₁) = 0 km/hr (standstill)
- Final velocity (v₂) = 40 km/hr
- Time (t) = 10 seconds

⚠️ Important: Units must be consistent. Velocity is in km/hr, but time is in seconds. We need to convert either velocity to m/s or time to hours.

✔ Let’s convert 40 km/hr → m/s:
> 40 km/hr = 40 × (1000 m / 3600 s) = 40 × (5/18) = 11.11 m/s

Now plug into formula:
> a = (v₂ – v₁) / t = (11.11 m/s – 0) / 10 s = 1.11 m/s²

✔ Answer: 1.11 m/s²

*(Note: The handwritten answer says “4 km/hr/s” — this is technically correct if you keep units in km/hr per second, but it’s not standard SI units. In physics, we usually use m/s².)*

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🔢 Problem 7:

> A car accelerates from 25 km/hr to 55 km/hr in 30 seconds. What is its acceleration?

Given:
- v₁ = 25 km/hr
- v₂ = 55 km/hr
- t = 30 s

First, convert velocities to m/s:

> 25 km/hr = 25 × (5/18) ≈ 6.94 m/s
> 55 km/hr = 55 × (5/18) ≈ 15.28 m/s

Now calculate acceleration:

> a = (15.28 – 6.94) / 30 ≈ 8.34 / 30 ≈ 0.278 m/s²

✔ Answer: 0.278 m/s²

*(Handwritten answer says “1 km/hr/s” — again, unit mismatch. 1 km/hr/s = 1000 m / 3600 s² ≈ 0.278 m/s² — so numerically correct, but unit-wise unconventional.)*

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🔢 Problem 8:

> A train is accelerating at a rate of 2.0 m/s². If its initial velocity is 20 m/s, what is its velocity after 30 seconds?

This is a rearranged version of the same formula. We now solve for final velocity:

> v₂ = v₁ + a × t

Given:
- v₁ = 20 m/s
- a = 2.0 m/s²
- t = 30 s

> v₂ = 20 + (2.0 × 30) = 20 + 60 = 80 m/s

✔ Answer: 80 m/s

*(Handwritten answer matches this.)*

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🔢 Problem 9:

> A runner achieves a velocity of 11.1 m/s after he begins. What distance did he cover? (Assume constant acceleration from rest.)

Wait — the question seems incomplete. It says:

> “A runner achieves a velocity of 11.1 m/s after he begins.” — but doesn’t specify time or acceleration.

However, looking at the handwritten solution, it appears they assumed constant acceleration from rest (v₁ = 0) and used the equation:

> d = (v₁ + v₂)/2 × t

But t is not given. Looking at the handwritten work:

They wrote:
> d = (0 + 11.1)/2 × 4.5 = 5.55 × 4.5 = 24.975 m ≈ 25 m

So they assumed time = 4.5 seconds — perhaps from context or prior problem? Since it’s not stated in the printed question, this is an assumption.

Alternatively, if we assume constant acceleration and want to find distance without time, we can use:

> v₂² = v₁² + 2ad

But we still need a or t.

Since the handwritten solution uses t = 4.5 s, and gets d = 25 m, we’ll go with that — assuming the missing info was meant to be provided.

✔ Answer: 25 meters (assuming t = 4.5 s)

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## ✔ Final Answers Summary:

| Problem | Question | Answer |
|---------|--------------------------------------------------------------------------|--------------------|
| 6 | Car from 0 to 40 km/hr in 10 s | 1.11 m/s² |
| 7 | Car from 25 to 55 km/hr in 30 s | 0.278 m/s² |
| 8 | Train at 2.0 m/s², starts at 20 m/s, after 30 s | 80 m/s |
| 9 | Runner reaches 11.1 m/s — distance covered? (assume t=4.5s) | 25 m |

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## 📌 Key Concepts:

- Acceleration = change in velocity over time → a = Δv / t
- Always check units — convert km/hr to m/s if needed (× 5/18)
- For distance under constant acceleration:
- d = v_avg × t = (v₁ + v₂)/2 × t
- Or d = v₁t + ½at²
- Or v₂² = v₁² + 2ad

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Let me know if you’d like to see any part explained further!