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Question
a honey bee travels at a constant speed. it takes a honey bee 6 seconds to fly 48 yards. let d be the distance the honey bee travels in yards, and t represent the time in seconds. now look at the second equation. what does $t = \frac{1}{8}d$ tell you about the relationship between the distance the bee can fly, d, and the time it takes the bee to fly, t? you can replace the number of yards, d, with 1 yard in your equations and solve for t to figure out how many seconds it take the honey bee to fly 1 yard. $d = 8t$ $t = \frac{1}{8}d$ it takes the honey bee $\frac{1}{8}$ second to fly 1 yard. the honey bee can fly $\frac{1}{8}$ yard in 1 second.
Step1: Analyze the equation \( t = \frac{1}{8}d \)
The equation \( t = \frac{1}{8}d \) is a linear relationship between time \( t \) (in seconds) and distance \( d \) (in yards). To understand the time per yard, we can substitute \( d = 1 \) yard into the equation.
Step2: Substitute \( d = 1 \) into the equation
When \( d = 1 \), we have \( t=\frac{1}{8}\times1=\frac{1}{8} \) seconds. This means that for each yard the honey bee flies (when \( d = 1 \)), the time \( t \) taken is \( \frac{1}{8} \) seconds. So the equation \( t=\frac{1}{8}d \) tells us that the time it takes the honey bee to fly (in seconds) is \( \frac{1}{8} \) of the distance it flies (in yards), or equivalently, it takes the honey bee \( \frac{1}{8} \) second to fly 1 yard.
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The equation \( t = \frac{1}{8}d \) shows that the time \( t \) (in seconds) the honey bee takes to fly is \( \frac{1}{8} \) of the distance \( d \) (in yards) it flies. Specifically, it takes the honey bee \( \frac{1}{8} \) second to fly 1 yard (since when \( d = 1 \), \( t=\frac{1}{8} \) second), and as the distance \( d \) increases, the time \( t \) increases proportionally with a constant of proportionality \( \frac{1}{8} \) (meaning time is directly proportional to distance with a rate of \( \frac{1}{8} \) seconds per yard).