Question

Sonya drives 170 miles at a certain speed. After stopping at a rest stop, she drives an additional 320 miles at a speed 5 mph slower than before the stop. If she drove 3 hours longer after the stop than before the stop, what was her speed before the stop?

Answer 1

45.75 mph

Explanation:

Let's call Sonya's speed before the rest stop "x" (in miles per hour). Then, we know that her speed after the rest stop was "x-5" miles per hour.

We can use the formula: distance = speed x time, to set up two equations based on the given information:

Equation 1: 170 = x * t1 (where t1 is the time Sonya drove before the rest stop)

Equation 2: 320 = (x-5) * (t1+3) (where t1+3 is the time Sonya drove after the rest stop)

We can solve for t1 in Equation 1 by dividing both sides by x:

t1 = 170/x

Now we can substitute this expression for t1 into Equation 2 and simplify:

320 = (x-5) * (170/x + 3)

320 = 170(x-5)/x + 3(x-5)

Multiplying both sides by x gives:

320x = 170(x-5) + 3x(x-5)

320x = 170x - 850 + 3x^2 - 15x

Simplifying and rearranging terms gives a quadratic equation:

3x^2 - 165x + 850 = 0

We can solve for x using the quadratic formula:

x = [165 ± sqrt(165^2 - 4(3)(850))] / (2*3)

x = [165 ± sqrt(27225 - 10200)] / 6

x = [165 ± sqrt(17025)] / 6

x = [165 ± 130.5] / 6

x = 45.75 or x = 9.25

We can ignore the solution x = 9.25 since it doesn't make sense in the context of the problem (Sonya's speed cannot be negative). Therefore, Sonya's speed before the rest stop was 45.75 miles per hour