Question 258233: two cars drive the 120 miles from town A to town B. The first car goes 10 mph faster than the second, and completes the journey in 1 hour less time. Let s be the speed of the slower car.
(A) In terms of s, what is the speed of the faster car?
(B) In terms of S, how long does the slower car take to complete the trip?
(C) In terms of S, how long does the faster car take to complete the trip?
(D) Express the conditions "the faster car completes the journey in one hour less than the slower car" as an equation involving s using the answers to part b and c.
(E) Solve the equation in part d to find the speed of the slower car.
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! (A) In terms of s, what is the speed of the faster car?
faster car travels at s + 10 miles per hour.
(B) In terms of S, how long does the slower car take to complete the trip?
rate * time = distance.
this makes time = distance / rate.
slower car takes (120/s) hours to complete the trip.
(C) In terms of s, how long does the faster car take to complete the trip?
faster car takes (120 / (s+10)) hours to complete the trip.
(D) Express the conditions "the faster car completes the journey in one hour less than the slower car" as an equation involving s using the answers to part b and c.
the time it takes the slower car is expressed by the equation:
h[s] = 120/s where h represents the time in hours.
the time it takes the faster car is expressed by the equation:
h[f] = 120/(s+10) where h represents the time in hours.
the statement that it takes the faster car one less hour to complete the trip can be expressed by the equation h[f] = h[s] - 1.
replace h[f] and h[s] with their respective equivalents and you get:
120/(s+10) = 120/s - 1
(E) Solve the equation in part d to find the speed of the slower car.
the equation in part D is:
120/(s+10) = 120/s - 1
multiply both sides of this equation by s * (s+10) to get:
120 * s = 120 * (s+10) - (1 * (s+10) * s)
simplify to get:
120 * s = 120 * s + 1200 - (s^2 + 10*s))
simplify further to get:
120 * s = 120 * s + 1200 - s^2 - 10*s
subtract 120 * s from both sides of this equation to get:
0 = 120 * s - 120 * s + 1200 - s^2 - 10*s
combine like terms to get:
0 = 1200 - s^2 - 10*s
add s^2 to both sides of this equation and add 10*s to both sides of this equation and subtract 1200 from both sides of this equation to get:
s^2 + 10*s - 1200 = 0
factor this quadratic equation to get:
(s+40) * (s-30) = 0
solve for s to get s = -40 or s = 30
since s can't be negative, then s must equal 30.
speed of the slower car is 30 miles per hour.
speed of the faster car is 40 miles per hour.
since time = distance / rate, then:
the time it takes the slower car = 120/30 = 4 hours.
the time it takes the faster car = 120/40 = 3 hours.
the faster car takes 1 hour less than the slower car so the equation is satisfied.
answer to part E is that the speed of the slower car is 30 miles per hour.
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