Physics worksheet on calculating acceleration using formulas and numerical examples.

Let’s solve each problem step by step. We’ll use the formula for acceleration:

a = (v - u) / t

Where:
- a = acceleration
- v = final velocity
- u = initial velocity
- t = time

We must make sure all units match before calculating. If they don’t, we convert them first.

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Problem 1:

Initial Velocity (u) = 0 m/s
Final Velocity (v) = 24 m/s
Time (t) = 3 s

a = (24 - 0) / 3 = 24 / 3 = 8 m/s²

✔ Correct as shown in image.

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Problem 2:

u = 0 m/s
v = 35 m/s
t = 5 s

a = (35 - 0) / 5 = 35 / 5 = 7 m/s²

✔ Correct as shown in image.

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Problem 3:

u = 20 km/h
v = 60 km/h
t = 10 s → But velocities are in km/h, time in seconds → Need to convert!

Convert km/h to m/s:
Multiply by 1000/3600 = 5/18

So:
u = 20 × (5/18) = 100/18 ≈ 5.56 m/s
v = 60 × (5/18) = 300/18 ≈ 16.67 m/s

Now calculate:

a = (16.67 - 5.56) / 10 ≈ 11.11 / 10 ≈ 1.11 m/s²

But wait — the student wrote “4 km/h/s” — that’s also a valid unit if you keep everything in km/h and seconds.

Let’s try that way:

a = (60 - 20) km/h / 10 s = 40 / 10 = 4 km/h per second

That’s acceptable too — it means every second, speed increases by 4 km/h.

✔ So both answers are correct depending on units. The student used km/h/s — which is fine.

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Problem 4:

u = 50 mi/h
v = 150 mi/h
t = 2 min → Convert minutes to hours? Or keep consistent?

If we want acceleration in mi/h per minute, then:

a = (150 - 50) mi/h / 2 min = 100 / 2 = 50 mi/h per minute

✔ Matches student answer.

Alternatively, if converting to standard units (mi/h²), but since question doesn’t specify, this is acceptable.

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Problem 5:

u = 25 km/h
v = 1200 km/h
t = 2 min

Again, let’s do it in km/h per minute:

a = (1200 - 25) / 2 = 1175 / 2 = 587.5 km/h per minute

✔ Student wrote 587.5 km/h/min — correct.

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Problem 6:

Car accelerates from standstill (0 km/h) to 60 km/h in 10 seconds.

Find acceleration in km/h/s.

a = (60 - 0) / 10 = 6 km/h per second

✔ Student got this right.

Note: They also converted to m/s² later — let’s check that too.

60 km/h = 60 × (1000/3600) = 60 × (5/18) = 300/18 ≈ 16.67 m/s

a = (16.67 - 0) / 10 = 1.667 m/s² — which matches their calculation.

Good job!

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Problem 7:

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

First, find change in velocity: 55 - 25 = 30 km/h

Time = 30 seconds

Acceleration = 30 km/h / 30 s = 1 km/h per second

✔ Student got this right.

They also converted to m/s²:

25 km/h = 25 × 5/18 ≈ 6.94 m/s
55 km/h = 55 × 5/18 ≈ 15.28 m/s
Change = 15.28 - 6.94 = 8.34 m/s
a = 8.34 / 30 ≈ 0.278 m/s² — which matches their work.

Perfect.

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Problem 8:

Train accelerating at 2.0 m/s². Initial velocity = 0. What is velocity after 30 seconds?

Use: v = u + a×t
v = 0 + 2.0 × 30 = 60 m/s

Student did:
v = 2.0 m/s² × 30 s = 60 m/s — then converted to km/h:

60 m/s × (3600/1000) = 60 × 3.6 = 216 km/h

✔ All correct.

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Problem 9:

Runner accelerates at 11 m/s² for 5 seconds. Starts from rest.

Distance covered?

Use: d = ut + ½at²
u = 0, so d = ½ × 11 × (5)² = 0.5 × 11 × 25 = 5.5 × 25 = 137.5 meters

Wait — student wrote 45.8 m? That’s wrong.

Let me recalculate:

d = ½ × a × t²
= 0.5 × 11 × 25
= 5.5 × 25
= 137.5 m

Also, final velocity: v = u + at = 0 + 11×5 = 55 m/s — student said 12.5 m/s? That’s incorrect.

Where did they get 12.5? Maybe they did 11 × 5 / something? No.

Actually, looking at their writing:
“v = 11 m/s² × 5s = 55 m/s” — that part is correct.
Then they say “= 12.5 m/s”? That seems like a typo or miswrite.

And distance: they wrote d = ½(11)(5)^2 = 45.8 — but 0.5×11×25 = 137.5, not 45.8.

Maybe they mistakenly used t=3? 0.5×11×9 = 49.5 — still not 45.8.

Or maybe they used average velocity incorrectly?

Average velocity = (initial + final)/2 = (0 + 55)/2 = 27.5 m/s
Distance = avg vel × time = 27.5 × 5 = 137.5 m — same result.

So student’s answer of 45.8 m is incorrect.

Correct answer: 137.5 meters

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## Final Answers Summary:

1. 8 m/s²
2. 7 m/s²
3. 4 km/h/s (or ~1.11 m/s²)
4. 50 mi/h/min
5. 587.5 km/h/min
6. 6 km/h/s (or ~1.67 m/s²)
7. 1 km/h/s (or ~0.278 m/s²)
8. 60 m/s (or 216 km/h)
9. 137.5 meters ← corrected from student’s 45.8 m

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Final Answer:
For problem 9, the correct distance the runner covers is 137.5 meters.