A child in an airport is able to cover 380 meters in 4 minutes running at a steady speed down a?

Question

moving sidewalk in the direction of the sidewalk's motion. Running at the same speed in the direction opposite to the sidewalk's movement, the child is able to cover 392 meters in 8 minutes. What is the child's running speed on a still sidewalk, and what is the speed of the moving sidewalk?

Answer 1

Simple algebra...

c = child's speed in meter/minute (on steady ground)
w = walkway speed in m/min

c + w = 380/4 = 95 m/min
c - w = 392/8 = 49 m/min

Adding these together:
2c = 144 m/min
c = 72 m/min

Then w can be found with either of these equations:
w = 95 - 72 = 23 m/min.
w = 72 - 49 = 23 m/min.

So the child walks normally at 72 m/min.
And the walkway moves at 23 m/min.

In kilometers per hour:
72 * 60 / 1000 = 4.32 k/h
23 * 60 / 1000 = 1.38 k/h

You didn't specify desired units for speed but either answer is correct.

Answer 2

the speed of the kid (k) plus the walkway (w) is 380/4 = 95 m/min
the speed of the kid minus the walkway is 392/8 = 49 m/min
the system is
k+w=95
k-w=49
add them to get
2k=144
k=72
72+w=95
w=23
and that's in meters per minute

Answer 3

380 in 4 minutes is 5.7 km/hr (380*60/4)
392 in 8 minutes is 2.94 km/hr (392*60/8)

Speed of child + Speed of moving sidewalk is
(a) Sc+Sms= 5.7
(b) Sc-Sms= 2.94

[(a)+(b)]/2 = Sc = 4.32 km/hr (Speed of child on a still sidewalk)

Sms=5.7-Sc = 1.38 km/hr

Answer 4

Can you please rephrase your question?