Physics worksheet on acceleration calculations, including formulas and practice problems.

Let’s go through each problem one by one. We’ll use the formula for acceleration:

a = (v_f - v_i) / t

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

We’ll calculate each answer carefully, checking units and math.

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Problem 1:

Initial Velocity = 0 km/hr
Final Velocity = 24 km/hr
Time = 3 s

a = (24 - 0) / 3 = 8 km/hr/s

✔ Correct as written.

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Problem 2:

Initial Velocity = 0 m/s
Final Velocity = 35 m/s
Time = 5 s

a = (35 - 0) / 5 = 7 m/s² (or m/s/s)

✔ Correct as written.

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Problem 3:

Initial Velocity = 20 km/hr
Final Velocity = 60 km/hr
Time = 10 s

a = (60 - 20) / 10 = 40 / 10 = 4 km/hr/s

✔ Correct as written.

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Problem 4:

Initial Velocity = 50 m/s
Final Velocity = 150 m/s
Time = 5 s

a = (150 - 50) / 5 = 100 / 5 = 20 m/s²

✔ Correct as written.

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Problem 5:

Initial Velocity = 25 km/hr
Final Velocity = 1200 km/hr
Time = 2 min → but we need to be careful! The unit of time is minutes, so acceleration will be in km/hr per minute, which is unusual but acceptable if specified.

a = (1200 - 25) / 2 = 1175 / 2 = 587.5 km/hr/min

The student wrote “588” — that’s rounded. But let’s check: 1175 ÷ 2 = 587.5 → so it should be 587.5, not 588 unless rounding is allowed.

⚠️ Minor error: Should be 587.5 km/hr/min, not 588.

But since the worksheet shows 588, maybe they rounded up. We’ll note this.

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Problem 6:

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

a = (60 - 0) / 10 = 6 km/hr/s

✔ Correct as written.

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Problem 7:

Car accelerates from 25 km/hr to 55 km/hr in 30 seconds.

a = (55 - 25) / 30 = 30 / 30 = 1 km/hr/s

✔ Correct as written.

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Problem 8:

Train accelerating at 2.0 km/hr/s. Initial velocity = 20 km/hr. Time = 30 seconds. Find final velocity.

Use:
a = (v_f - v_i) / t

Rearrange to solve for v_f:
v_f = a × t + v_i

Plug in:
v_f = 2.0 km/hr/s × 30 s + 20 km/hr
= 60 km/hr + 20 km/hr = 80 km/hr

✔ Correct as written.

Note: Units work out because “s” cancels with “/s”, leaving km/hr.

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Problem 9:

Runner achieves velocity of 11.1 m/s after 9 seconds (starts from rest).

Part A: Acceleration?

a = (11.1 - 0) / 9 = 1.233... ≈ 1.23 m/s²

Student wrote 1.23 m/s² — correct.

Part B: Distance covered?

Use: d = ½ × a × t²

d = 0.5 × 1.23 × (9)²
First, 9² = 81
Then, 0.5 × 1.23 = 0.615
Then, 0.615 × 81 = ?

Let’s compute:

0.615 × 80 = 49.2
0.615 × 1 = 0.615
Total = 49.2 + 0.615 = 49.815 m

Student wrote 49.8 m — that’s correctly rounded.

Also, alternative method: average speed × time

Average speed = (initial + final)/2 = (0 + 11.1)/2 = 5.55 m/s
Distance = 5.55 × 9 = 49.95 m → wait, that’s different!

Wait — inconsistency here.

If acceleration is constant, then yes, average speed = (v_i + v_f)/2.

But if a = 11.1 / 9 = 1.2333... m/s² exactly, then:

d = ½ × a × t² = 0.5 × (11.1/9) × 81
= 0.5 × 11.1 × 9
= 0.5 × 99.9 = 49.95 m

Ah! So there’s a rounding issue.

Student used a = 1.23 m/s² (rounded), then d = 0.5 × 1.23 × 81 = 49.815 ≈ 49.8 m

But if we use exact value:

a = 11.1 / 9 = 1.2333... m/s²
d = 0.5 × (11.1/9) × 81 = 0.5 × 11.1 × 9 = 49.95 m

So which is correct?

Actually, the problem says “achieves a velocity of 11.1 m/s after 9 s”. That implies 11.1 is given, so we should use it directly.

Better approach: Use d = (v_i + v_f)/2 × t

d = (0 + 11.1)/2 × 9 = 5.55 × 9 = 49.95 m

Or using d = ½ a t² with a = 11.1/9:

d = 0.5 × (11.1/9) × 81 = 0.5 × 11.1 × 9 = 49.95 m

So the more accurate answer is 49.95 m, which rounds to 50.0 m if using 3 significant figures.

But student got 49.8 m due to rounding acceleration early.

This is a common mistake — don’t round intermediate values!

However, since the problem gives 11.1 (which has 3 sig figs), and 9 s (possibly exact), we can keep 3 sig figs.

49.95 m → rounds to 50.0 m

But student wrote 49.8 m — which comes from using a = 1.23 instead of 1.233...

So technically, 49.8 m is slightly inaccurate.

But perhaps the worksheet expects them to use the rounded acceleration.

Looking back at the student’s work:

They wrote:
a = 11.1 / 9 = 1.23 m/s² (they rounded too early)
then d = 0.5 × 1.23 × 81 = 49.8 m

To avoid error, better to do:

d = ½ × (11.1 / 9) × 81 = ½ × 11.1 × 9 = 49.95 m

So correct distance is 49.95 m, or 50.0 m with proper rounding.

But since the student’s answer is based on their calculation path, and many teachers accept it, we’ll note the discrepancy.

For accuracy, we should say 49.95 m.

But let’s see what the problem asks: “What distance did he cover?”

Given data: v_f = 11.1 m/s, t = 9 s, starts from rest.

Best formula: d = average velocity × time = (0 + 11.1)/2 × 9 = 5.55 × 9 = 49.95 m

So Final Answer for distance: 49.95 m

But student wrote 49.8 m — close but not exact.

In educational context, sometimes they expect you to use the rounded acceleration. However, to be precise, we’ll go with 49.95 m.

Alternatively, if we use a = 11.1 / 9 = 1.2333..., then d = 0.5 * 1.2333... * 81 = let's calculate:

1.2333... * 81 = 100 (exactly? 11.1/9 * 81 = 11.1 * 9 = 99.9, then half is 49.95)

Yes.

So I think the most accurate answer is 49.95 m.

But since the student might have been instructed to round acceleration first, and the worksheet shows 49.8, we’ll mention both.

For now, in our solution, we’ll provide the accurate value.

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Now, compiling all answers with corrections where needed:

1. 8 km/hr/s ✔
2. 7 m/s² ✔
3. 4 km/hr/s ✔
4. 20 m/s² ✔
5. 587.5 km/hr/min (student wrote 588 — minor rounding)
6. 6 km/hr/s ✔
7. 1 km/hr/s ✔
8. 80 km/hr ✔
9. Acceleration: 1.23 m/s² (if rounded), but exactly 1.233...; Distance: 49.95 m (not 49.8)

But since the question is to solve accurately, we’ll give precise answers.

However, looking at the format, the student filled in blanks, and we are to verify or correct.

But the instruction is: "Solve the problem accurately."

So we’ll list all answers correctly.

Also, for problem 5, time is 2 min, so unit is km/hr per minute — that’s fine.

Final Answers:

1. 8 km/hr/s
2. 7 m/s²
3. 4 km/hr/s
4. 20 m/s²
5. 587.5 km/hr/min
6. 6 km/hr/s
7. 1 km/hr/s
8. 80 km/hr
9. Acceleration: 1.23 m/s² (with 3 sig figs, since 11.1 has 3, 9 may be exact) — actually 11.1 / 9 = 1.233..., so 1.23 m/s² is acceptable with 3 sig figs.
Distance: Using d = ½ a t² with a=1.2333... gives 49.95 m, which to 3 sig figs is 50.0 m.
Or using d = (v_avg)*t = 5.55 * 9 = 49.95 m → 50.0 m.

But 11.1 has 3 sig figs, 9 is likely exact (counted), so distance should have 3 sig figs: 50.0 m.

Student wrote 49.8 m — which is incorrect due to premature rounding.

So for accuracy, we'll use 50.0 m.

But let's confirm calculation:

v_f = 11.1 m/s, t = 9 s, u = 0

a = Δv/t = 11.1 / 9 = 1.2333... m/s²

d = ut + ½at² = 0 + 0.5 * (11.1/9) * 81 = 0.5 * 11.1 * 9 = 0.5 * 99.9 = 49.95 m

49.95 rounded to 3 significant figures is 50.0 m (since 49.95 is closer to 50.0 than to 49.9).

Significant figures: 49.95 has four, but input 11.1 has three, so output should have three.

49.95 → to three sig figs: look at the third digit is 9, fourth is 5, so round up: 50.0 m.

Yes.

So distance is 50.0 m.

Student's 49.8 m is wrong.

Therefore, corrected answers:

Problem 5: 587.5 km/hr/min (not 588)
Problem 9 distance: 50.0 m (not 49.8)

Others are correct.

Now, for the final answer section, since the user didn't specify which problem, but the image has multiple, I think we need to provide all answers as solved.

But the instruction says "solve the problem", and there are multiple problems.

Perhaps we should list all answers.

Since it's a worksheet, and the student filled some, we can provide the correct versions.

But to follow the format, I'll provide the final answers for each numbered item.

Final Answer:

1. 8 km/hr/s
2. 7 m/s²
3. 4 km/hr/s
4. 20 m/s²
5. 587.5 km/hr/min
6. 6 km/hr/s
7. 1 km/hr/s
8. 80 km/hr
9. Acceleration: 1.23 m/s², Distance: 50.0 m

Note: For problem 9, acceleration is often reported as 1.23 m/s² (rounded from 1.233...), and distance as 50.0 m.

Some might write acceleration as 1.23 m/s² and distance as 49.95 m, but with sig figs, 50.0 m is better.

I think for consistency, we'll go with:

After careful calculation:

- Problem 5: (1200 - 25) / 2 = 1175 / 2 = 587.5 km/hr/min
- Problem 9 distance: 0.5 * (11.1 / 9) * 81 = 49.95 m → rounded to 50.0 m for 3 significant figures.

So final answers are as above.