Calculus Challenge Question URGENT!

Question

I need to catch my flight. I’m at the airport and I’m running late. I need to make it 1000m to get to my gate. I have two choices. I can either run or take the moving sidewalk. The moving sidewalk moves at a constant speed of 2m/s. If I’m running, I start out quick, but because I’m old I quickly tire and slow down.

Equations for these situations are:

M(t) = 2t

R(t) = √2000x (2000x = radicand)

I cannot run on the moving sidewalk because it’s too crowded, but I can to go onto or off of the moving sidewalk whenever I want to. Assume that I never recover my energy, so I can only have one burst of running.

1. How can I get to my gate in less than 500s? Explain the process and determine the minimum time it

takes me to reach the gate.

2. If the moving sidewalk moved at 2.5m/s, what is the best strategy and how quickly can I reach the gate?

3. If the moving sidewalk is moving at 2m/s but running speed is modelled by R(t) = 15³√x²(3rd root), what is the best strategy and how long will it take me to reach the gate?

Answer 1

I know this is 3 days ago but here is my solution.

Total distance is 1000 meters. Distance on Sidewalk is x, so distance running is (1000 - x).

time = distance/rate

The time running is (2m/sec)(t sec) = (1000 - x)m ⇒ t = (1000 - x)m/(2 m/s) = [500 - x/2] seconds.

The time on the Sidewalk is (2000t)^1/2 = x meters ⇒ [20√(5t)]m/s = x meters ⇒ t = (x²)/(2000) sec

Total time T = R(t) + M(t) = 2t + (2000t)^1/2

T = 500 - x/2 + (x²)/2000

Note that if x = 0 (spends all time running, T = 500 seconds.

If x = 1000 (spends all time on sidewalk), T = 500 seconds.

D_xT = -1/2 + 2x/1000 = x/500 - 1/2

Solving for x when the equation equals 0

x/500 - 1/2 = 0 ⇒ x = 250 meters.

Thus, distance on the sidewalk = 250 meters and distance running = 750 meters. Solving for time, we get

Running: 2t = 750, t = 375 seconds.

Sidewalk: √(2000t) = 250, t = 36.25 seconds

Total time is 406.25 seconds.

I'll leave it to you to do parts 2 and 3.