Question

cu 3-141. (from lesson 3.3.2)
the great frog-jumping contest is ready to start. ooma
and cruz have placed their frogs facing in the same
direction at 0, and when the bell rings, they both let go.
ooma’s frog jumps 3 feet, then 2 feet, gives one last hop
of 1 foot, and then stops. cruz’s frog turns completely
around, and hops in the opposite direction! cruz’s frog
takes a leap of 6 feet, followed by a jump of 4 feet, and
one final hop of 2 feet, and then stops.
a. represent each frog’s hops on the number line.
number line image with 0 in the middle
b. who should win the jumping contest? justify your
answer.

Explanation:

Part a (Representing on Number Line)

Step1: Ooma's Frog Movement

Ooma’s frog starts at \( 0 \), jumps \( 3 \) feet right (positive direction), then \( 2 \) feet right, then \( 1 \) foot right. So the positions are \( 0 \to 0 + 3 = 3 \to 3 + 2 = 5 \to 5 + 1 = 6 \). Mark these hops: from \( 0 \) to \( 3 \), \( 3 \) to \( 5 \), \( 5 \) to \( 6 \) on the number line (right of \( 0 \)).

Step2: Cruz's Frog Movement

Cruz’s frog starts at \( 0 \), jumps \( 6 \) feet left (negative direction, since it turns around), then \( 4 \) feet left, then \( 2 \) feet left. So positions: \( 0 \to 0 - 6 = -6 \to -6 - 4 = -10 \to -10 - 2 = -12 \). Mark these hops: from \( 0 \) to \( -6 \), \( -6 \) to \( -10 \), \( -10 \) to \( -12 \) on the number line (left of \( 0 \)).

Part b (Determining Winner)

Step1: Calculate Ooma's Final Position

Sum Ooma’s hops: \( 3 + 2 + 1 = 6 \) (positive, right of \( 0 \)).

Step2: Calculate Cruz's Final Position

Sum Cruz’s hops (magnitude, since direction is opposite, but we consider distance from start? Wait, no—wait, the problem says "jumping contest"—maybe total distance? Wait, no, maybe final position. Wait, Ooma’s frog is at \( 6 \), Cruz’s at \( -12 \). But distance from \( 0 \): Ooma is \( 6 \) units, Cruz is \( 12 \) units? Wait, no, wait—Cruz’s frog jumps in opposite direction, so each hop is negative. Wait, maybe the contest is about how far they move from start, regardless of direction? Wait, no, the problem says "jumping contest"—maybe total distance jumped? Wait, Ooma’s total distance: \( 3 + 2 + 1 = 6 \). Cruz’s total distance: \( 6 + 4 + 2 = 12 \). But that seems odd. Wait, no—wait, the problem says "facing in the same direction at 0", then Cruz’s frog turns around. Wait, maybe the final position: Ooma is at \( 6 \) (right), Cruz at \( -12 \) (left). But "win"—maybe who is farther from \( 0 \)? Cruz’s frog is at \( -12 \), which is \( 12 \) units from \( 0 \); Ooma at \( 6 \), \( 6 \) units. But that seems counterintuitive. Wait, maybe I misread. Wait, Ooma’s frog: 3, 2, 1 (all positive). Cruz’s frog: turns around, so hops are negative? Wait, no—"turns completely around, and hops in the opposite direction"—so initial direction was, say, positive, then Cruz’s frog goes negative. So each hop is negative. So Cruz’s hops: \( -6, -4, -2 \). Sum: \( -6 -4 -2 = -12 \). Ooma: \( 3 + 2 + 1 = 6 \). Now, if the contest is about how far they moved (total distance), Cruz’s total distance is \( 6 + 4 + 2 = 12 \), Ooma’s is \( 6 \). But maybe the problem is about final position—who is farther from \( 0 \)? Cruz is at \( -12 \), Ooma at \( 6 \). So Cruz’s frog is farther (12 units vs 6 units). But that seems odd. Wait, maybe the problem is about total distance jumped (regardless of direction). Let's check:

Ooma’s total distance: \( 3 + 2 + 1 = 6 \) feet.

Cruz’s total distance: \( 6 + 4 + 2 = 12 \) feet.

Since \( 12 > 6 \), Cruz’s frog jumped more total distance, so Cruz should win? Wait, but the problem says "jumping contest"—maybe total distance. Alternatively, maybe final position: Ooma at \( 6 \), Cruz at \( -12 \). The absolute value of Cruz’s position is \( 12 \), Ooma’s is \( 6 \). So Cruz’s frog is farther from the start, so Cruz wins.

Answer:

(Part b):
Cruz should win. Ooma’s frog ends at \( 6 \) (total distance \( 6 \) feet), Cruz’s frog ends at \( -12 \) (total distance \( 12 \) feet, or absolute position \( 12 \) units from \( 0 \)). Since \( 12 > 6 \), Cruz’s frog jumped a greater total distance (or is farther from the start), so Cruz wins.