<PRE><B>Question 258233</B>
(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 &quot;the faster car completes the journey in one hour less than the slower car&quot; 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&#39;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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