Question

examine the graph at the right.
two model helicopters take off at the same time. helicopter a rises 5 meters every 2 seconds. helicopter b rises 4 meters every 3 seconds.
a) fill in the missing ordered pairs on the coordinate graph at the right for each helicopter.
b) does the rise of each helicopter represent a proportional relationship? explain.
c) using ratios, explain why helicopter a is rising faster than helicopter b.
d) how much faster does helicopter a move than helicopter b?

Explanation:

Step1: Find speed of Helicopter A

Speed of A: $\frac{5}{2} = 2.5$ m/s

Step2: Find speed of Helicopter B

Speed of B: $\frac{4}{3} \approx 1.33$ m/s

Step3: Generate ordered pairs for A

For A: (0,0), (2,5), (4,10), (6,15)

Step4: Generate ordered pairs for B

For B: (0,0), (3,4), (6,8), (9,12)

Step5: Analyze proportional relationship

Check $\frac{\text{height}}{\text{time}}$ ratio consistency

Step6: Compare speeds with ratios

Simplify $\frac{5}{2}$ and $\frac{4}{3}$ to compare

Step7: Calculate speed difference

Subtract B's speed from A's speed

Answer:

a) Helicopter A ordered pairs: (0, 0), (2, 5), (4, 10), (6, 15)
Helicopter B ordered pairs: (0, 0), (3, 4), (6, 8), (9, 12)

b) Yes, each helicopter's rise represents a proportional relationship. For each helicopter, the ratio of height risen to time elapsed is constant (Helicopter A: $\frac{5}{2}$; Helicopter B: $\frac{4}{3}$), and the relationship passes through the origin (0,0), which defines a proportional relationship.

c) The ratio of height to time for Helicopter A is $\frac{5}{2} = 2.5$, and for Helicopter B it is $\frac{4}{3} \approx 1.33$. Since $2.5 > 1.33$, Helicopter A's height-to-time ratio is larger, meaning it rises faster.

d) Helicopter A moves $\frac{5}{2} - \frac{4}{3} = \frac{15}{6} - \frac{8}{6} = \frac{7}{6} \approx 1.17$ meters per second faster than Helicopter B.