Question

Determine the distance in feet that light can travel in vacuum during \(1.00 \mathrm{~ns}\).

Short answer

Answer: Approximately 0.98 feet.

Step-by-step solution

Convert the speed of light to feet per second

Since the speed of light is given in meters per second, we need to convert it to feet per second. To do this, we will use the conversion factor: \(1 \mathrm{~m} = 3.2808 \mathrm{~ft}\). So, we have: \(c_{ft} = c_{m} \times 3.2808\) \(c_{ft} = (3.00\times10^8) \mathrm{~(m/s) } \times 3.2808 \mathrm{~(ft/m)}\) \(c_{ft} = 9.8425 \times 10^8 \mathrm{~ft/s}\)

Convert the time from nanoseconds to seconds

Now we need to convert the given time, 1.00 ns, into seconds by using the conversion factor: \(1 \mathrm{~s} = 10^9 \mathrm{~ns}\). \(t_s = \dfrac{1.00 \mathrm{~ns}}{10^9}\) \(t_s = 1.00\times10^{-9} \mathrm{~s}\)

Calculate the distance using the speed of light and time

Now that we have the speed of light in feet per second and the time in seconds, we can use the equation: \(distance= speed \times time\) In our case: \(distance = c_{ft} \times t_s\) \(distance = (9.8425\times10^8 \mathrm{~ft/s}) (1.0\times10^{-9} \mathrm{~s})\) \(distance = 9.8425 \times 10^{-1} \mathrm{~ft}\)

Write down the final answer

The distance that light can travel in a vacuum during 1.00 ns is: \(distance = 9.8425 \times 10^{-1} \mathrm{~ft}\) It can be approximated to: \(distance \approx 0.98 \mathrm{~ft}\)

Key concepts

Unit Conversion

Unit conversion is a fundamental skill in solving physics problems. It involves changing a measurement from one unit to another without altering the value. This comes in handy when measurements are given in units that are not immediately usable in equations.
For example, the speed of light is often given in meters per second (m/s), but in certain situations, such as the exercise above, it might be more practical to use feet per second (ft/s). The conversion factor between meters and feet is:

• 1 meter = 3.2808 feet

To convert the speed of light from meters per second to feet per second, we multiply the speed by this conversion factor:

\[ c_{ft} = c_{m} \times 3.2808 \]
This step ensures that we have a consistent measurement system fit for our calculations. The converted speed was found to be approximately 9.8425 x 10^8 ft/s, which greatly facilitates solving the problem at hand.

Nanoseconds

Nanoseconds are an incredibly small unit of time, and understanding them is key for tackling high-speed calculations like those involving the speed of light. A nanosecond is one billionth of a second, symbolized by 'ns'. This means:

• 1 second = 1,000,000,000 nanoseconds
• 1 nanosecond = \(10^{-9}\) seconds

Converting from nanoseconds to seconds is essential because most equations use seconds as the standard unit of time. Thus, in our exercise, the conversion of 1.00 nanoseconds to seconds is done by dividing by one billion. This conversion gives us:

\[ t_s = \dfrac{1.00 \, \text{ns}}{10^9} = 1.00 \times 10^{-9} \, \text{s} \]
By converting to seconds, we align our units with the speed given and can proceed with calculations accurately.

Distance Calculation

Calculating distance when provided with speed and time is a straightforward process once unit conversions are completed. The fundamental equation used is:

\[ \text{distance} = \text{speed} \times \text{time} \]
In this context, after converting the speed of light to feet per second and time to seconds, we can determine how far light travels in a given time interval. Using the values calculated:
- Speed of light in feet per second: 9.8425 x 10^8 ft/s
- Time: 1.00 x 10^{-9} seconds

The calculation becomes:
\[ \text{distance} = (9.8425 \times 10^8 \, \text{ft/s}) \times (1.0 \times 10^{-9} \, \text{s}) \]
This results in a distance of 9.8425 x 10^{-1} feet, or approximately 0.98 feet. This exercise demonstrates the often surprising efficiency of light travel over extremely short periods, illustrating the breathtaking speed of light in the context of very brief time intervals like a nanosecond.