Greg Drove at a constant speed - Formulas and Applications

Question

Greg drove at a constant speed in a rainstorm for

220 miles. He took a​ break, and the rain stopped He then drove

106 miles at a speed that was 9 miles per hour faster than his previous speed. If he drove for 7 hours, find the​ car's speed for each part of the trip.

Answer 1

Let "s_i" denote velocity in units "mi/hr." for trips i, and "x_i" denote the hours traveled per trip.

Based on the problem, there are a total of two trips. It is also known that the velocity traveled during trip 2 is 7 more than the velocity traveled during trip 1 (*). The key to solve the problem is to create an algebraic expression or equation per trip.

According to (*), s₂ = s₁+9

(1) For trip 1, 220 miles = s₁*x₁

(2) For trip 2, 106 miles = s₂*x₂ = (s₁+9)*x₂

(3) For entire trip, x₁ + x₂ = 7

Adding equation (1) and (2) yields...326= s₁*x₁ + (s₁+9)*x₂ = s₁*x₁ + s₁*x₂ + 9*x₂ = s₁*(x₁ + x₂) + 9*x₂

Applying relation (3) by substitution yields....s₁*(7) + 9*x₂ = 326 --> s₁*(7) + 9*(7-x₁) = 326-->

s₁*(7) + 63- 9x₁ =326.

Since x₁ = 220/s₁ via (1), we have s₁*(7) + 63- 9(220/s₁) = 326.

Therefore, (7s₁² -1980)/s₁ = 326 - 63 --> (7s₁² -1980) = 263*s₁ or 7s₁² - 263*s -1980 = 0

Solving for s₁ yields....(263 +/- (263² -4*7*(-1980))^1/2 /(2*7) --> (263 + 353)/(14) = s₁ = 44 mph.

Therefore, substituting in (*) yields.. s₂ = s₁ +9 = 44 + 9 = 53 mph.

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Check:

(1) 220 = 44*x₁ --> x₁ = 220/44 = 5 hours.

(2) 106 = 53*x₂ --> x₂ = 106/53 = 2 hours.

(3) x₁ + x₂ = 7--> 5 + 2 = 7 ✔