Math Anyone?

Question

A child in an airport is able to cover 294 meters in 3 minutes running at a steady speed down a moving sidewalk in the direction of the sidewalk’s motion. Running at the same speed in the direction of the opposite to the sidewalk’s movement, the child is able to cover 240 meters in 5 minutes. What is the child’s running speed on a still sidewalk, and what is the speed of the moving sidewalk?

Answer 1

Ok, so convert minutes to second

For first sidewalk moving

294 meters in 180 sec

Second sidewalk(not moving)

240 meters in 300 sec then we use ratios to figure how much he ran in 180 sec by mutliply 240m by (180/300)

so dist in 180 sec is 144 meters

So first sidewalk do distance=velocity X time

294 metrs=180 sec X velocity

Velocity for 1=1.6333 m/s (combined with sidewalk speed)

Now for the second
144 meters=180 sec X velocity

Velocity for 2=.8 m/s (child's actual velocity -sidewalk's velocity)

If we add both velocities and divide by 2 we receive the child's velocity which is 1.21667 m/s.

Now we solve velocity 2 so .8m/s=1.21667m/s-sidewalk velocity

Childs velocity=1.21667 m/s and sidewalk velocity=.41667 m/s

Then we

Answer 2

Speed of child is 25 m/min (0.42 m/s) and the speed of the sidewalk is 73 m/min (1.2 m/s)

Quickly describing how i got the answer:
Traveling the same direction, the velocity of the sidewalk plus the velocity of the child must equal 294/3, or 98 m/min. Traveling against the sidewalk, the difference of the two velocities must equal 240/5, or 48m/min. Setting up a simultaneous equation, and solving...i get those answers

Answer 3

I came up with the following two equations:

walking speed + moving sidewalk = walking with sidewalk
walking speed - moving sidewalk = walking against sidewalk

Substitute the known values.

walking speed + moving sidewalk = 294m/3mins (98m/min)
walking speed - moving sidewalk = 240m/5mins (48m/min)

Add the two equations together to solve for walking speed.

2 * walking speed = 146m/min (solve for walking speed)
walking speed = 73m/min
moving sidewalk = 25m/min (sub in one of the original equations)

The child running on a still sidewalk would be 73m/min. The moving sidewalk is moving 25m/min.

It's been awhile since I have done these type of equations. How did I do?

Answer 4

consider x is speed of child
and y is speed of sidewalk

so in first case -- both ar in same direction so relative speed is
x+y = 294/3 = 98

In second case both are in opposite direction so relative speed is
x-y = 240/5 = 48

so if we add both
2x = 98+48 = 146
x= 73
y = 98-73 = 25

So child speed is 73 m/sec
and sidewalk speed is 25 m/sec

Answer 5

child 1.21 m/s
sidewalk 0.41 m/s

say vc is child and vs is sidewalk
294m / 3 min is 1.63m/s
240m / 5 min is .8 m/s

vc+vs=1.63
second equation is either vc-vs=0.8 or vs-vc=0.8 but the child is moving forward when is is running in the direction of the opposite to the sidewalk’s movement so vc > vs
so second equation is vc-vs=.8

solving that system should not be a problem =)

Answer 6

Childs velocity=1.21667 m/s
sidewalk velocity=.41667 m/s

I calculate on my brain

Answer 7

I agree w/ Techman20