a) [tex]v=\frac{d_{tot}}{t_{tot}}=\frac{(3 h)(60 mph)+20 mi}{3 h +t_2}[/tex]
The average speed is equal to the ratio between the total distance ([tex]d_{tot}[/tex] and the total time taken ([tex]t_{tot}[/tex]):
[tex]v=\frac{d_{tot}}{t_{tot}}[/tex]
the distance travelled by the trucker in the first 3 hour can be written as the time multiplied by the velocity:
[tex]d_1 = (3 h)(60 mph)=180 mi[/tex]
So the total distance is
[tex]d_{tot}=d_1 +d_2 = 180 mi+20 mi=200 mi[/tex]
The total time is equal to the first 3 hours + the time taken to cover the following 20 miles in the city:
[tex]t_{tot}=3 h +t_2[/tex]
So, the equation can be rewritten as:
[tex]v=\frac{d_{tot}}{t_{tot}}=\frac{(3 h)(60 mph)+20 mi}{3 h +t_2}[/tex]
b) 0.50 h (half a hour)
Since we know the value of the average speed, [tex]v=57.14 mph[/tex], we can substitute it into the previous equation to find the value of [tex]t_2[/tex], the time the trucker drove in the city:
[tex]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[/tex]
