SOLUTION: Speed velocity worksheet 2 - Studypool - Free Printable

Let's solve each problem from Unit 1: Uniform Motion – Worksheet 8, step by step, with explanations.

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1. What is the average speed of a cheetah that sprints 100 m in 4 s? How about if it sprints 50 m in 2 s?

Formula for average speed:
$$
\text{Average Speed} = \frac{\text{Distance}}{\text{Time}}
$$

- For 100 m in 4 s:
$$
\frac{100\,\text{m}}{4\,\text{s}} = 25\,\text{m/s}
$$

- For 50 m in 2 s:
$$
\frac{50\,\text{m}}{2\,\text{s}} = 25\,\text{m/s}
$$

✔ Answer: Both cases give an average speed of 25 m/s.

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2. If a car moves with an average speed of 60 km/hr for an hour, it will travel a distance of 60 km. How far will it travel if it continues this average rate for 4 hrs?

Use:
$$
\text{Distance} = \text{Speed} \times \text{Time}
$$

$$
60\,\text{km/hr} \times 4\,\text{hr} = 240\,\text{km}
$$

✔ Answer: 240 km

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3. A runner makes one lap around a 200 m track in a time of 25.0 s. What was the runner's average speed?

$$
\text{Average Speed} = \frac{200\,\text{m}}{25.0\,\text{s}} = 8.0\,\text{m/s}
$$

✔ Answer: 8.0 m/s (as given)

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4. Light and radio waves travel through a vacuum at a speed of nearly $3.00 \times 10^8$ m/s. How far is a light year (the distance light travels in a year)?

We need to compute:
$$
\text{Distance} = \text{Speed} \times \text{Time}
$$

First, find how many seconds in a year:

- 1 minute = 60 s
- 1 hour = 60 min = 3600 s
- 1 day = 24 hr = $24 \times 3600 = 86,400$ s
- 1 year ≈ 365.25 days (accounting for leap years)

So:
$$
\text{Seconds in a year} = 365.25 \times 86,400 = 31,557,600\,\text{s} \approx 3.15576 \times 10^7\,\text{s}
$$

Now:
$$
\text{Distance} = (3.00 \times 10^8\,\text{m/s}) \times (3.15576 \times 10^7\,\text{s}) = 9.467 \times 10^{15}\,\text{m}
$$

Rounded to three significant figures:
$$
\boxed{9.50 \times 10^{15}\,\text{m}}
$$

✔ Answer: $9.50 \times 10^{15}$ m

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5. A motorist travels 406 km during a 7.0 hr period. What was the average speed in km/hr and m/s?

#### In km/hr:
$$
\frac{406\,\text{km}}{7.0\,\text{hr}} = 58\,\text{km/hr}
$$

#### Convert to m/s:

- 1 km = 1000 m
- 1 hr = 3600 s

So:
$$
58\,\text{km/hr} = \frac{58 \times 1000}{3600} = \frac{58000}{3600} \approx 16.11\,\text{m/s} \approx 16\,\text{m/s}
$$

✔ Answer: 58 km/hr, 16 m/s

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6. A bullet is shot from a rifle with a speed of 720 m/s. What time is required for the bullet to strike a target 3240 m away?

Use:
$$
\text{Time} = \frac{\text{Distance}}{\text{Speed}} = \frac{3240\,\text{m}}{720\,\text{m/s}} = 4.5\,\text{s}
$$

✔ Answer: 4.5 s

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7. Light from the sun reaches the earth in 8.3 minutes. The speed of light is $3.0 \times 10^8$ m/s. In kilometers, how far is the Earth from the Sun?

Convert time to seconds:
$$
8.3\,\text{min} = 8.3 \times 60 = 498\,\text{s}
$$

Now:
$$
\text{Distance} = (3.0 \times 10^8\,\text{m/s}) \times 498\,\text{s} = 1.494 \times 10^{11}\,\text{m}
$$

Convert meters to kilometers:
$$
1.494 \times 10^{11}\,\text{m} = 1.494 \times 10^8\,\text{km} \approx 1.5 \times 10^8\,\text{km}
$$

✔ Answer: $1.5 \times 10^8$ km

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8. An auto travels at a rate of 25 km/hr for 4 minutes, then at 50 km/hr for 8 minutes, and finally at 20 km/hr for 2 minutes. Find the total distance covered in km and the average speed for the complete trip in m/s.

#### Step 1: Convert times to hours

- 4 min = $ \frac{4}{60} = \frac{1}{15} $ hr
- 8 min = $ \frac{8}{60} = \frac{2}{15} $ hr
- 2 min = $ \frac{2}{60} = \frac{1}{30} $ hr

#### Step 2: Compute distances

- First leg: $ 25 \times \frac{1}{15} = \frac{25}{15} = 1.666... \approx 1.67 $ km
- Second leg: $ 50 \times \frac{2}{15} = \frac{100}{15} = 6.666... \approx 6.67 $ km
- Third leg: $ 20 \times \frac{1}{30} = \frac{20}{30} = 0.666... \approx 0.67 $ km

Total distance:
$$
1.67 + 6.67 + 0.67 = 9.01 \approx 9.0\,\text{km}
$$

(But exact: $ \frac{25}{15} + \frac{100}{15} + \frac{20}{30} = \frac{125}{15} + \frac{2}{6} = 8.\overline{3} + 0.666... = 9.0 $ km)

✔ Total distance: 9.0 km

#### Total time:
$ 4 + 8 + 2 = 14 $ minutes = $ \frac{14}{60} = \frac{7}{30} $ hr ≈ 0.2333 hr

#### Average speed in km/hr:
$$
\frac{9.0\,\text{km}}{0.2333\,\text{hr}} = 38.57\,\text{km/hr}
$$

Convert to m/s:
$$
38.57\,\text{km/hr} = \frac{38.57 \times 1000}{3600} = \frac{38570}{3600} \approx 10.71\,\text{m/s} \approx 10.7\,\text{m/s}
$$

✔ Answer: 9 km, 10.7 m/s

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9. If you traveled one mile at 100 mph and another mile at 1 mph, what would be your average speed?

Important: Average speed is total distance / total time, not the arithmetic mean.

- Distance: 2 miles

- Time for first mile: $ \frac{1}{100} $ hr = 0.01 hr
- Time for second mile: $ \frac{1}{1} $ hr = 1 hr
- Total time = 0.01 + 1 = 1.01 hr

$$
\text{Average speed} = \frac{2\,\text{mi}}{1.01\,\text{hr}} \approx 1.98\,\text{mph} \approx 1.98\,\text{mph}
$$

Wait — but the answer says 1.98 mph? Let’s check.

Actually:
$$
\frac{2}{1.01} = 1.980198... \approx 1.98\,\text{mph}
$$

But the hint says "answer: 1.98 mph" — yes, correct.

✔ Answer: 1.98 mph

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10. What is your average speed in each case?

#### a) Run 100 m at 5.0 m/s, then walk 100 m at 1.0 m/s

- Time for running: $ \frac{100}{5.0} = 20 $ s
- Time for walking: $ \frac{100}{1.0} = 100 $ s
- Total distance = 200 m
- Total time = 120 s
- Average speed = $ \frac{200}{120} = 1.67\,\text{m/s} $

But answer says 1.7 m/s — rounded to two sig figs → ✔ 1.7 m/s

#### b) Run for 100 s at 5.0 m/s, then walk for 100 s at 1.0 m/s

- Distance running: $ 5.0 \times 100 = 500 $ m
- Distance walking: $ 1.0 \times 100 = 100 $ m
- Total distance = 600 m
- Total time = 200 s
- Average speed = $ \frac{600}{200} = 3.0\,\text{m/s} $

✔ Answer: a) 1.7 m/s, b) 3.0 m/s

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11. A race car driver must average 200 km/hr for four laps to qualify. He averages only 170 km/hr over the first two laps. What average speed must he maintain for the last two laps?

Let each lap be distance $ d $. So total distance = $ 4d $

To qualify:
$$
\text{Average speed} = \frac{4d}{T_{\text{total}}} = 200\,\text{km/hr}
\Rightarrow T_{\text{total}} = \frac{4d}{200} = 0.02d\,\text{hr}
$$

Time for first two laps:
$$
T_1 = \frac{2d}{170} = \frac{2d}{170} = 0.0117647d\,\text{hr}
$$

Remaining time:
$$
T_2 = 0.02d - 0.0117647d = 0.0082353d\,\text{hr}
$$

Distance remaining: $ 2d $

Required speed:
$$
v = \frac{2d}{0.0082353d} = \frac{2}{0.0082353} \approx 242.7\,\text{km/hr}
$$

✔ Answer: Approximately 243 km/hr

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12. A car traveling 90 km/hr is 100 m behind a truck traveling 50 km/hr. How long will it take the car to reach the truck?

Relative speed = $ 90 - 50 = 40\,\text{km/hr} $

Convert to m/s:
$$
40\,\text{km/hr} = \frac{40 \times 1000}{3600} = \frac{40000}{3600} \approx 11.11\,\text{m/s}
$$

Distance to cover = 100 m

Time:
$$
t = \frac{100\,\text{m}}{11.11\,\text{m/s}} \approx 9.0\,\text{s}
$$

✔ Answer: 9.0 seconds

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13. The peregrine falcon dives at 97.2 m/s vertically. From a height of 100 m, how much time does this give a rabbit below to react?

Assuming constant velocity (no acceleration), use:
$$
\text{Time} = \frac{\text{Distance}}{\text{Speed}} = \frac{100\,\text{m}}{97.2\,\text{m/s}} \approx 1.0288\,\text{s} \approx 1.03\,\text{s}
$$

✔ Answer: 1.03 seconds

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✔ Summary of Answers:

| Problem | Answer |
|--------|--------|
| 1 | 25 m/s both times |
| 2 | 240 km |
| 3 | 8.0 m/s |
| 4 | $9.50 \times 10^{15}$ m |
| 5 | 58 km/hr, 16 m/s |
| 6 | 4.5 s |
| 7 | $1.5 \times 10^8$ km |
| 8 | 9 km, 10.7 m/s |
| 9 | 1.98 mph |
| 10a | 1.7 m/s |
| 10b | 3.0 m/s |
| 11 | ~243 km/hr |
| 12 | 9.0 s |
| 13 | 1.03 s |

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