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Chapter 3: Problem 8

Salmon speed. You have designed a special circuit to measure the swimming speed of a salmon. The circuit has a length of \(62.8 \mathrm{~m}\). You measure the time a salmon takes to swim one lap to be \(20.6 \mathrm{~s}\). (a) What is the swimming speed of a salmon? Your assistant insists that you would get better precision if you instead measured the time the salmon took to swim 10 rounds. You find that the salmon uses \(206.0 \mathrm{~s}\) to swim 10 rounds. (b) What is the speed of the salmon? (c) Does this produce better accuracy? Can you give other examples of situations where this strategy would improve the accuracy?

Short Answer

Expert verified
a) 3.05 m/s. b) 3.05 m/s. c) Yes, measuring over ten laps provides better precision.

Step by step solution

01

Calculate the speed for one lap

The formula for speed is \(v = \frac{d}{t}\), where \(d\) is the distance and \(t\) is the time taken. Here, \(d = 62.8\) meters and \(t = 20.6\) seconds. Substituting the values into the formula, we get:\[ v = \frac{62.8}{20.6} \approx 3.05 \mathrm{m/s} \]
02

Calculate the speed for ten laps

Using the same formula \(v = \frac{d}{t}\), but this time, \(d\) is the distance of ten laps which is \(10 \times 62.8 = 628\) meters and \(t = 206.0\) seconds. Substituting the values, we get:\[ v = \frac{628}{206.0} \approx 3.05 \mathrm{m/s} \]
03

Determine if this produces better accuracy

Both methods give the same speed of approximately \(3.05 \mathrm{m/s}\). However, measuring the time over a longer distance (10 laps) averages out more minor timing inaccuracies, providing better precision. Other examples include measuring the speed of a car over multiple kilometers or the flow rate of water over a more extended period.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Speed Calculation
To find the swimming speed of a salmon, we use the formula for speed: \( v = \frac{d}{t} \). Here, \( d \) is the distance traveled and \( t \) is the time taken. For the salmon, one lap of the circuit is \( 62.8 \) meters, and the time taken is \( 20.6 \) seconds. Substituting these values gives: \[ v = \frac{62.8}{20.6} \approx 3.05 \mathrm{m/s} \] This shows that the speed of the salmon is approximately \( 3.05 \) meters per second. Use the same approach if the distance or time changes. It's a universal formula for speed.
Accuracy Improvement
Improving the accuracy of speed measurements involves reducing potential errors. By measuring the salmon over ten laps instead of just one, minor timing inaccuracies have less impact. For ten laps, the distance becomes \( 10 \times 62.8 = 628 \) meters, and the time is \[ 206.0 \] seconds. Calculating the speed: \[ v = \frac{628}{206.0} \approx 3.05 \mathrm{m/s} \] Even small timing errors get averaged out over longer distances. The result is a more precise measurement.
Measurement Techniques
Different techniques can measure speed more accurately. Here, we used timing gates to track how long the salmon took to swim a circuit. When using timing devices:
  • Ensure devices are calibrated correctly.
  • Prefer longer measurement periods to reduce relative timing errors.
  • Repeat measurements several times and calculate the average.
For more accuracy, avoid human error by using automated systems.
Distance and Time Relationship
The relationship between distance and time is fundamental in calculating speed. Speed is directly proportional to the distance traveled and inversely proportional to the time taken.
Mathematically, it’s shown as: \[ v = \frac{d}{t} \] If you double the distance (keeping time constant), the speed is doubled. Similarly, if the time is halved (keeping distance constant), the speed doubles. Understanding this helps in designing better experiments for speed measurement.
More trials and extended distances average out timing inaccuracies, providing consistent and accurate speed values.

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Most popular questions from this chapter

Bacterial volume. A bacteria is like a cylinder with length \(4 \mu \mathrm{m}\), and radius \(1 \mu \mathrm{m}\). (a) Find its volume in \(\mu \mathrm{m}^{3}\). (b) Find its volume in \(m^{3}\). (c) Find its volume in liters.

Miles per hour. You car speedometer is showing both \(\mathrm{km} / \mathrm{h}\) and mph, miles per hour. I mile is \(1609.34 \mathrm{~m}\). (a) If your speedometer is showing \(70 \mathrm{~km} / \mathrm{h}\). What does it show in \(\mathrm{mph}\) ? (b) If your speedometer is showing \(55 \mathrm{mph}\). What does it show in \(\mathrm{km} / \mathrm{h}\) ?

Kilometers per hour. A car is driving at \(144 \mathrm{~km} / \mathrm{h}\). Find the velocity in \(\mathrm{m} / \mathrm{s}\).

Acceleration of gravity. The acceleration of gravity is approximately \(g=\) \(9.8 \mathrm{~m} / \mathrm{s}^{2}\) (a) Find the acceleration of gravity in feet per second squared, \(\mathrm{ft} / \mathrm{s}^{2} .1 \mathrm{foot}\) is \(0.3048 \mathrm{~m}\). (b) Find the acceleration of gravity in kilometers per hour squared, \(\mathrm{km} / \mathrm{h}^{2}\).

Sphere mass and volume. A small steel sphere has a radius of \(1.2 \mathrm{~mm}\). (a) What is the volume of the sphere? (b) The density of the particular steel alloy used is \(\rho=7782 \mathrm{~kg} / \mathrm{m}^{3}\). What is the mass of the sphere?
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