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Question
a leaf hangs from a branch 12 feet in the air. it falls to the ground at a rate of 0.25 feet per second. which graph could represent the leaf’s height in feet as a function of time, in seconds, after leaving the branch? three graphs of falling leaf with axes height (feet) and time (seconds) are shown, each with a line plot and radio buttons below
Step1: Define the height function
The initial height \( h_0 = 12 \) feet, and the rate of falling is \( r = 0.25 \) feet per second. So the height \( h(t) \) as a function of time \( t \) (in seconds) is \( h(t)=12 - 0.25t \). This is a linear function with a slope of \(- 0.25\) (negative because height decreases over time) and a y - intercept of 12.
Step2: Analyze the y - intercept
At \( t = 0 \) (when the leaf just starts to fall), the height should be 12 feet. So the graph should pass through the point \((0,12)\).
Step3: Analyze the slope
The slope of the line is \(-0.25\), which means for each second that passes, the height decreases by 0.25 feet. Let's find the time when the leaf reaches the ground (\( h(t)=0 \)):
But we can also analyze the steepness. A slope of \(- 0.25\) is a relatively gentle slope (less steep than a slope of - 1, for example).
Now let's analyze the three graphs:
- The first graph: At \( t = 0 \), the height is above 12 (since it's above the 12 - foot mark on the y - axis), so it does not have a y - intercept of 12. Eliminate the first graph.
- The second graph: It passes through \((0,12)\) (y - intercept is 12) and has a gentle slope (consistent with a slope of - 0.25), which matches our function \( h(t)=12 - 0.25t \).
- The third graph: At \( t = 0 \), the height is 15 feet (not 12 feet), so it does not match our initial condition. Eliminate the third graph.
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The second graph (the one with the title "Falling Leaf", y - intercept at (0,12) and a gentle decreasing slope)