Question

Sally travels by car from one city to another. She drives for 30.0 min at 80.0 km/hr, 12.0 min at 105.0 km/hr, and 45.0 min at 40.0 km/hr, and she spends 15.0 min eating lunch and buying gas.
A. Determine the average speed for the trip.
B. Determine the total distance traveled.

Answer 1

Total distance = (30/60 x 80) + (12/60 x 105) + (45/60 x 40) = 0.5 x 80 + 0.2 x 105 + 0.75 x 40 = 40 + 21 + 30 = 91 km

Average distance = total distance / time taken = 91 / (30/60 + 12/60 + 45/60) = 91/ (0.5 + 0.2 + 0.75) = 91/1.45 = 62.76 km/hr

Answer 2

Answer:

A) The average speed for the trip is $v_{average} =53,53\frac{km}{h}$.

B) The total distance traveled is $d_{total}=91\ km$.

Explanation:

Knowing that 1 hour has 60 minutes, we have that for each part of the trip:

                                                $v_{1}=80\frac{km}{h}\ t_{1}=0,5\ h$

                                                $v_{2}=105\frac{km}{h}\ t_{2}=0,2\ h$

                                                $v_{3}=40\frac{km}{h}\ t_{3}=0,75\ h$

                                                $v_{4}=0\frac{km}{h}\ t_{4}=0,25\ h$

Total distance traveled

The distance traveled is the speed multiplied by a period of time. The total distance traveled is calculated adding up the speeds multiplied by the given periods of time.

                                                  $d_{total}=d_{1}+d_{2}+d_{3}$

                                                  $d_{total}=v_{1}.t_{1}+v_{2}.t_{2}+v_{3}.t_{3}$

                                  $d_{total}=80\frac{km}{h}.0,5\ h+105\frac{km}{h}.0,2\ h+40\frac{km}{h}.0,75\ h$

                                                $d_{total}=40\ km+21\ km+30\ km$

                                                $d_{total}=91\ km$

B) The total distance traveled is $d_{total}=91\ km$.

Average speed

The average speed could be calculated using the total distance traveled $d_{total}$ divided by the total time $t_{total}$. We need to take into account the time spent eating lunch and buying gas.

                                                $v_{average}=\frac{d_{total}}{t_{total}}$

                                              $v_{average}=\frac{v_{1}.t_{1}+v_{2}.t_{2}+v_{3}.t_{3}+v_{4}.t_{4}}{t_{1}+t_{2}+t_{3}+t_{4}}$                                                           $v_{average}=\frac{91\ km}{0,5\ h+0,2\ h+0,75\ h+0,25\ h}$

                                                $v_{average}=\frac{91\ km}{1,70\ h}$

                                                $v_{average}=53,53\frac{km}{h}$

A) The average speed for the trip is $v_{average} =53,53\frac{km}{h}$.