john spent the first 100 miles of his road trip in traffic. When the traffic cleared, he was able to drive twice as fast for the remaining 400 miles. If the total trip took 10 hours how fast was he moving in traffic?
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Answer:
Step-by-step explanation:
Let x = speed for the first 96 miles
x - 9 = speed for the last 100 miles
Distance = (rate)(time), so time = (distance)/(rate)
(time for first part of trip) + (time for last part of trip) = 4 hours
So, (96)/(x) + (100)/(x - 9) = 4
Multiply both sides by the LCD (x)(x - 9) to get
96(x - 9) + 100x = 4x(x - 9)
96x - 864 +100x = 4x2 - 36x
4x2 - 232x + 864 = 0
Divide both sides by 4 to obtain x2 - 58x +216 = 0
(x - 4)(x - 54) = 0
x = 4 or x = 54
Note that x can't equal 4 because, in that case, x - 9 would be negative, so x = 54 miles per hour.
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Arthur D. answered • 12/27/14
TUTOR 4.9 (55)
Effective Mathematics Tutor
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distance=rate*time
d=rt
96=(r+9)(t)
100=(r)(4-t)
96/(r+9)=t from the first equation
substitute this into the second equation
100=(r)(4-[96/(r+9)])
100=4r-96r/(r+9)
multiply both sides by (r+9)
100(r+9)=4r(r+9)-(r+9)(96r)/(r+9)
100r+900=4r2+36r-96r
100r+900=4r2-60r
get all terms on one side of the equation
4r2-60r-100r-900=0
4r2-160r-900=0
divide both sides by 4
r2-40r-225=0
factor (225=3*3*5*5=(3*3*5)(5)=45*5 and 45-5=40)
(r-45)(r+5)=0
r-45=0
r=45 mph
r+9=45+9=54 mph for the first 96 miles