Question

When truckers are on long-haul drives, their driving logs must reflect their average speed. Average speed is the total distance driven divided by the total time spent driving. A trucker drove 3 hours on a freeway at 60 miles per hour and then drove 20 miles in the city. The trucker’s average speed was 57.14 miles per hour.
a. Write an equation that models the situation.
b. How long was the trucker driving in the city to the nearest hundredth of an hour?

Answer 1

a) $v=\frac{d_{tot}}{t_{tot}}=\frac{(3 h)(60 mph)+20 mi}{3 h +t_2}$

The average speed is equal to the ratio between the total distance ($d_{tot}$ and the total time taken ($t_{tot}$):

$v=\frac{d_{tot}}{t_{tot}}$

the distance travelled by the trucker in the first 3 hour can be written as the time multiplied by the velocity:

$d_1 = (3 h)(60 mph)=180 mi$

So the total distance is

$d_{tot}=d_1 +d_2 = 180 mi+20 mi=200 mi$

The total time is equal to the first 3 hours + the time taken to cover the following 20 miles in the city:

$t_{tot}=3 h +t_2$

So, the equation can be rewritten as:

$v=\frac{d_{tot}}{t_{tot}}=\frac{(3 h)(60 mph)+20 mi}{3 h +t_2}$

b) 0.50 h (half a hour)

Since we know the value of the average speed, $v=57.14 mph$, we can substitute it into the previous equation to find the value of $t_2$, the time the trucker drove in the city:

$v=\frac{200 mi}{3h +t_2}\\3h+t_2 = \frac{200 mi}{v}\\t_2 = \frac{200 mi}{v}-3h=\frac{200 mi}{57.14 mph}-3 h=0.50 h$