Accelerated maths

Question

Mitchell decided to get some extra exercise at the airport by walking to her departure gate at her normal pace, but on a lengthy moving walkway.

After going 100 meters (that is getting 100m closer to her departure gate), Mitchell dropped her passport onto the walkway. But she didn’t notice at first and continued walking towards her gate. After walking another 90 seconds, she finally realised that she had dropped her passport. She immediately turned around and jogged in the direction of the walkways movement. Mitchell’s jogging pace is exactly twice as fast as her walking pace. She caught up with her passport 10m from the start of the moving walkway.

how fast does the walkway move ?

Answer 1

Let's assume,

The speed of the walkway is x m/s ←

Mitchell's normal walking speed is y m/s →

Mitchell's normal jogging speed is 2y m/s →

Initially Mitchell started walking to the direction opposite to that of the walkways movement.

∴ Mitchell's effective speed is (y+x) m/s with respect to the walkway.

After dropping the passport, Mitchell covered a distance of 90(y+x) m w.r.t. the walkway.

Therefore, at the time she turned around, the distance between Mitchell and her passport is 90(y+x) m

By this time, the passport is (100-90x) m away from the start of the walkway.

At the time she turned around, she is (100 + 90y) m away from the start of the walkway.

[start] [passport] [Mitchell]

↓ ↓ ↓

|--------------- (100-90x) ----------- • ------ 90(y+x) ------- • --------

|---------------------------- (100+90y) ----------------------- • --------

Now, she turned around and started moving in the direction of the walkways movement at a speed of 2y m/s.

∴ Now, Mitchell's effective speed is (2y-x) m/s w.r.t. the walkway.

Mitchell caught up with her passport 10 m from the start of the walkway.

Therefore,

The passport had travelled [(100 - 90x) - 10] m at a speed of x m/s and

Mitchell had to cover [(100 + 90y) - 10] m at a speed of 2y m/s in the same amount time.

They both took the same amount of time to cover their respective distances w.r.t. to a stationary observer (the start of the walkway).

And this time was t = [(100 - 90x) - 10] / x = (90 - 90x) / x seconds



∴ t = [(100 - 90x) - 10] / x = [(100 + 90y) - 10] / 2y

⇒ (90 - 90x) / x = (90 + 90y) / 2y

⇒ (1 - x) / x = (1 + y) / 2y

⇒ (1 - x) ⋅ 2y = x ⋅ (1 + y)

⇒ 2y - 2xy = x + xy

⇒ x + 3xy - 2y = 0

⇒ x = 2y / (1 + 3y)