mmelissa102000
contestada

Two trains leave the station at the same time, one heading west and the other east. The westbound train travels 20 miles per hour faster than the eastbound train. If the two trains are 800 miles apart after 5 hours, what is the rate of the westbound train?

Respuesta :

jcherry99

Givens

W = E + 20


d = 800 miles

d1 = Westbound train.

d2 = Eastbound train.


t = 5 hours.


Equations

West bound

W*5 = d1 miles

d1 = W*5


East bound

E*5 = (800 - d1)


Step One

Substitute d1 into the Eastbound train.

E*5 = (800 - 5*W)


Step Two

Substitute E = W + 20 on the left side of the above equation.

(W + 20)5 = 800 - 5W


Step Three

Solve Remove the brackets

5W + 100 = 800 - 5W Add 5W to both sides.

10W + 100 = 800 Subtract 100 from both sides.

10W = 800 - 100

10W = 700

W = 700/10

W = 70


E = W + 20

E = 70 + 10 = 90


Check

70*5 = 350

90*5 = 450

Total distance = 800

Let's assume that the speed is positive towards east.


So, the eastbound train has speed [tex] v [/tex], while the other train has speed [tex] -(v+20)=-v-20 [/tex]. In fact, its speed is greater than the other train's, but in the opposite direction.


So, the eastbound train follows the rule:


[tex] p_e(t) = vt [/tex]


while the westbound train follows the rule


[tex] p_w(t) = (-v-20)t [/tex]


where [tex] p_(t)[/tex] is the position of the train, in miles after [tex] t [/tex] hours.


So, we know that


[tex] p_e(5) - p_w(5) = 800 [/tex]


which means


[tex] 5v - 5(-v-20) = 800 [/tex]


And we can easily rearrange and solve for [tex] v [/tex]:


[tex] 5v+5v+100=800 [/tex]


[tex] 10v=700 [/tex]


[tex] v=70 [/tex]


But this was the speed of the eastbound train. The speed of the other train was greater than this by [tex] 20 [/tex] units, and in the opposite direction, so it is [tex] 90 [/tex] miles per hour in the opposite direction, or [tex] -90 [/tex] mph

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