Question

If the cheetah made a round-trip and took half the amount of time on the return trip as on the front end of the trip, what would be the relationship between the average rates on each leg of the trip? Using a complete sentence, explain how you arrived at this conclusion.

Answer 1

Answer:

The average rate on second leg of the trip is twice of average rate on first leg of the trip.

Step-by-step explanation:

The formula for average speed is

$Speed=\frac{Distance}{Time}$

$s=\frac{d}{t}$

$s\propto\frac{1}{t}$

It means the speed is inversely proportional to the time. If the speed increases then time deceases and if speed decrees then time increases.

It is given that the cheetah made a round-trip and took half the amount of time on the return trip as on the front end of the trip.

Let s₁ and s₂ are average speed on first and second leg respectively. t₁ is time on the first trip.

$s_1=\frac{d}{t_1}$

$s_2=\frac{d}{\frac{t_1}{2}}$

$s_2=2\frac{d}{t_1}$

$s_2=2s_1$

Therefore, the average rate on second leg of the trip is twice of average rate on first leg of the trip.

Answer 2

The relationship between the average rates is that ; The average rate of return trip is twice the average rate of front trip.

Let :

• Time taken on front trip = t
• Time taken on return trip = half of t = 1/2 × t = t/2

Relationship between speed(Rate) and time :

• Rate α 1/time (Rate is inversely proportional to time)

For front trip:

• Rate, r = 1/t

For return trip :

• Rate = 1/(t÷2)
• Rate = 1 × 2/t
• Rate = 2/t

Comparing the rates :

• Rate of front trip = 1/t
• Rate of return trip = 2/t

• Hence, rate of return trip = 2 × (rate of front trip) = 2 × 1/t

The average rate of return trip is twice the average rate of front trip.

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