What’s the answer

Question

At the​ airport, Person A and Person B are walking at the same speed to catch their​ flight, but Person B decides to step onto the moving​ sidewalk, while Person A continues to walk on the stationary sidewalk. If the sidewalk moves at 1 meter per​ second, and it takes Person B 35 seconds less to walk the 210-meter ​distance, at what speed are Person A and Person B​ walking? 

Answer 1

Given:

1- At the beginning person A speed is same as person B. That is V_A = V_B

2 - Velocity of walking speed = 1m/s

3- It takes person B 35 seconds less to walk the 210 meters, t_B = t_A -35 s

Solution:

Find t_B knowing that the speed of the sidewalk is constant 1m/s meaning that acceleration is = 0

t_B= x / V_B , t_B = (210 m / 1 m/s) = 210 s

Find t_A , t_A = t_B+35 s = 210 s + 35 s = 245 s

Person A has a constant speed on the stationary sidewalk V_A = x / t_A, V_A = 210 m/ 245 s = 0.86 m/s

Please note that this solution is valid only if the walking speed of person B when he goes on the moving sidewalk is < the moving sidewalk speed, which is the assumption made and proven that person A and person B speeds at the beginning are equal to 0.86 m/s < the sidewalk speed of 1 m/s

Answer 2

The first sentence states that both A and B are walking at the same speed.

That speed is 1 mps