Youngkwon C. answered 02/26/16
Tutor
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Knowledgeable and patient tutor with a Ph.D. in Electrical Eng.
Hi Anna,
If we denote the normal speed as v
and the travel time with the speed of v as t,
we can express
v·t = 200 (Eq. 1)
With the speed increased by 10 m/h,
the train can travel 200 miles in (t - 1) hours.
That is,
(v + 10)·(t - 1) = 200
v·t + 10t - v - 10 = 200
Plugging v·t = 200 into the equation right above,
we get
10t - v = 10 (Eq. 2)
From Eq. 2, we get
v = 10t - 10 (Eq. 3)
Plugging Eq. 3 into Eq. 1,
(10t - 10)·t = 200
10t2 - 10t - 200 = 0
t2 - t - 20 = 0
(t - 5)·(t + 4) = 0
t = 5 or t = -4
As t represents travel time,
only t = 5 can be a meaningful solution.
Plugging t = 5 into Eq. 1,
v·t = v·5 = 200
v = 40
So, the normal speed v of the train is 40 miles per hour.
Hope this helps.