SOLUTION: The Gare Montparnasse train station in Paris has a high-speed version of a moving walkway. If he walks while riding this moving walkway, Jean Claude can travel 200 meters in 30 sec

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Question 839898: The Gare Montparnasse train station in Paris has a high-speed version of a moving walkway. If he walks while riding this moving walkway, Jean Claude can travel 200 meters in 30 seconds less time than if he stands still on the moving walkway. If Jean Claude walks at a normal rate of 1.5 meters per second, what is the speed of the Gare Montparnasse walkway?
Found 3 solutions by richwmiller, ikleyn, josgarithmetic:
Answer by richwmiller(17219) About Me  (Show Source):
You can put this solution on YOUR website!
r*t=d
(1.5+y)(t-30)=200,
y*t=200
t = 80, y = 5/2 =2.5 meters per second walkway

Answer by ikleyn(52803) About Me  (Show Source):
You can put this solution on YOUR website!
.
The Gare Montparnasse train station in Paris has a high-speed version of a moving walkway.
If he walks while riding this moving walkway, Jean Claude can travel 200 meters in 30 seconds
less time than if he stands still on the moving walkway. If Jean Claude walks at a normal rate
of 1.5 meters per second, what is the speed of the Gare Montparnasse walkway?
~~~~~~~~~~~~~~~~~~~ 

Let x be the speed of the walkway, in meters per second.


When Jean Claude stands still on the walkway, he actually moves relatively to the floor
at the speed of x m/second.


The time to travel 200 meters at this speed is  200%2Fx  seconds.


When Jean Claude walks on the walkway with his relative speed of 1.5 m/s relative to the walkway,
he actually moves relatively to the floor at the speed of (1.5+x) m/second.


The time to travel 200 meters at this speed is  200%2F%28x%2B1.5%29  seconds.


The difference of the two travel times is 30 seconds - so, we write this time equation

    200%2Fx - 200%2F%28x%2B1.5%29 = 30  seconds.    (1)


This is your setup equation.
At this point, the setup is complete, and now your task is to solve it.


To solve the equation, multiply both sides by x*(x+1.5).  You will get

   200*(x+1.5) - 200*x = 30x*(x+1.5).


Simplify step by step

    200x + 300 - 200x = 30x^2 + 45x,

    300 = 30x^2 + 45x,


divide all the terms by 15

    20 = 2x^2 + 3x,

    2x^2 + 3x - 20 = 0.


Apply the quadratic formula

    x%5B1%2C2%5D = %28-3+%2B-+sqrt%283%5E2+-4%2A2%2A%28-20%29%29%29%2F%282%2A2%29 = %28-3+%2B-+sqrt%28169%29%29%2F4 = %28-3+%2B-+13%29%2F4.


You want the positive root  x = %28-3+%2B+13%29%2F2 = 10%2F4 = 2.5 m/s.


It is the speed of the walkway.    ANSWER


CHECK.  Check the validity of equation (1):  200%2F2.5 - 200%2F%281.5%2B2.5%29 = 200%2F2.5 - 200%2F4 = 30.0 seconds.

        Which is precisely correct !

At this point, the problem is solved completely.



Answer by josgarithmetic(39620) About Me  (Show Source):
You can put this solution on YOUR website!
Person normally walks 1.5 m/s
w, the speed of the walkway on its own 
                  speed         time in seconds       distance in meters

WITH WALKAY        w+1.5        200/(w+1.5)             200

ONLY WALKWAY        w           200/w                   200

DIFFERENCE                      30

highlight_green%28200%2Fw-200%2F%28w%2B1.5%29=30%29

Algebraic steps will lead to w%5E2%2B1.5w-10=0.

General solution and choosing the positive part is w=%28-1.5%2Bsqrt%281.5%5E2%2B4%2A10%29%29%2F2

w=%286.5-1.5%29%2F2

highlight%28w=2.5%29