Question

name:
passing on the trail
there is a hiking trail near the town where han and jada live that starts at a parking lot and ends at a lake. han and jada both decide to hike from the parking lot to the lake and back, but they start their hikes at different times.
at the time that han reaches the lake and starts to turn back, jada is 0.6 miles away from the parking lot and hiking at a constant speed of 3.2 miles per hour toward the lake. han’s distance, d, from the parking lot can be expressed as ( d = -2.4t + 4.8 ), where t represents the time in hours since he left the lake.

1. what is an equation for jada’s distance from the parking lot as she heads toward the lake?
2. draw both graphs: one representing han’s equation and one representing jada’s equation. it is important to be very precise.

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Explanation:

Question 1

Step1: Determine Han's lake distance

Han's equation is \( d = -2.4t + 4.8 \). When he reaches the lake, \( d = 0 \) (distance from parking lot to lake is total distance, so when returning, starting from lake, \( d = 0 \) at lake). Solve \( 0 = -2.4t + 4.8 \).
\( 2.4t = 4.8 \) → \( t = 2 \) hours. But at the time Han starts returning (reaches lake), Jada is 0.6 miles from parking lot, hiking toward lake at 3.2 mph. Wait, actually, the total distance from parking lot to lake: when Han starts (t=0), his distance from parking lot is 4.8 miles (since \( d(0) = 4.8 \)), so that's the one-way distance. So Jada is hiking toward the lake, so her distance \( d \) from parking lot should be a linear equation with positive slope (since she's moving toward lake, distance from parking lot increases? Wait no: wait, parking lot to lake is 4.8 miles. Jada is hiking toward the lake, so her distance from parking lot starts at some initial distance and increases? Wait no, if she's hiking toward the lake, starting from parking lot? Wait no, the problem says "Jada is 0.6 miles away from the parking lot and hiking at a constant speed of 3.2 miles per hour toward the lake". Wait, no: when Han reaches the lake (starts to turn back), Jada is 0.6 miles from parking lot, hiking toward the lake. Wait, maybe I misread. Let's re-express:

Han's equation: \( d = -2.4t + 4.8 \). This is distance from parking lot as he hikes (starts at parking lot? No, wait, when t=0, d=4.8, so he starts at the lake (d=4.8 is lake, d=0 is parking lot), and is hiking back to parking lot (hence negative slope). So the lake is 4.8 miles from parking lot.

At the time Han reaches the lake (d=0), solve \( 0 = -2.4t + 4.8 \) → \( t = 2 \) hours. At that time (t=2), Jada is 0.6 miles from parking lot, hiking toward the lake (so her distance from parking lot will increase? Wait no, if she's hiking toward the lake, starting from parking lot, her distance from parking lot would increase. Wait, parking lot to lake is 4.8 miles. So if she's at 0.6 miles from parking lot, hiking toward lake, her distance from parking lot is \( d = 0.6 + 3.2(t - 2) \), because at t=2, she is at 0.6 miles, and then moves toward lake (increasing distance from parking lot) at 3.2 mph. Wait, but maybe the time when Han starts (t=0) is when Jada also starts? No, the problem says "they start their hikes at different times". Wait, maybe Han's equation is when he starts hiking (from lake) at t=0, moving toward parking lot (so d decreases from 4.8 to 0 over t=2 hours). At the time Han reaches the lake (t=2), Jada is 0.6 miles from parking lot, hiking toward lake (so her distance from parking lot is \( d = 0.6 + 3.2(t - 2) \))? Wait, no, maybe the total distance from parking lot to lake is 4.8 miles (from Han's equation, when t=0, d=4.8, so he is at the lake, hiking back to parking lot). So Jada is hiking from parking lot toward the lake, so her distance from parking lot is \( d = v t + d_0 \), but at t=2 (when Han reaches lake), she is at 0.6 miles. Wait, maybe I got the direction wrong. Let's re-express:

Han's motion: starts at lake (d=4.8) at t=0, moves toward parking lot (d decreases) at 2.4 mph (since slope is -2.4), so time to reach parking lot is \( 4.8 / 2.4 = 2 \) hours (correct, as before).

Jada's motion: starts at some time, but at t=2 (when Han reaches parking lot? No, Han reaches lake at t=0? Wait no, Han's equation: d is distance from parking lot. So when t=0, d=4.8: he is at the lake (4.8 miles from parking lot), hiking back to parking lot (so d decreases as t increases). So when he reaches the parking l…

Step1: Graph Han's equation (\( d = -2.4t + 4.8 \))

• This is a linear equation with slope -2.4 and y-intercept 4.8.
• When \( t = 0 \), \( d = 4.8 \) (plot (0, 4.8)).
• When \( d = 0 \), \( 0 = -2.4t + 4.8 \) → \( t = 2 \) (plot (2, 0)).
• Draw a line through these two points.

Step2: Graph Jada's equation (\( d = 3.2t + 0.6 \))

• Linear equation with slope 3.2 and y-intercept 0.6.
• When \( t = 0 \), \( d = 0.6 \) (plot (0, 0.6)).
• When \( d = 4.8 \) (lake), \( 4.8 = 3.2t + 0.6 \) → \( 3.2t = 4.2 \) → \( t = 1.3125 \) (plot (1.3125, 4.8)).
• Draw a line through (0, 0.6) and (1.3125, 4.8).

Answer:

\( d = 3.2t + 0.6 \)

Question 2