Question

Driving along a crowded freeway, you notice that it takes a time tt to go from one mile marker to the next. When you increase your speed by 7.4 mi/hmi/h , the time to go one mile decreases by 15 ss . What was your original speed?

Answer 1

Answer:

38.6 mi/h

Explanation:

7.4 mi/h = 7.4mi/h * (1/60)hour/min * (1/60) min/s = 0.00206 mi/s

Let v (mi/s) be your original speed, then the time t it takes to go 1 mi/s is

t = 1/v

Since you increase v by 0.00206 mi/s, your time decreases by 15 s, this means

t - 15 = 1/(v+0.00206)

We can substitute t = 1/v to solve for v

$\frac{1}{v} - 15 = \frac{1}{v + 0.00206}$

We can multiply both sides of the equation with v(v+0.00206)

v+0.00206 - 15v(v+0.00206) = v

$-15v^2 - 0.0308v + 0.00206 = 0$

$v= \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

$v= \frac{0.03083\pm \sqrt{(-0.03083)^2 - 4*(-15)*(0.00205)}}{2*(-15)}$

$v= \frac{0.03083\pm0.35}{-30}$

v = -0.01278 or v = 0.01 0724 mi/s

Since v can only be positive we will pick v = 0.010724 mi/s or 0.010724*3600 = 38.6 mi/h