1 a. David's airplane trip took 3.8 hours. For one-half of that time, the airplane flew at a speed
of 680 miles per hour, and for the rest of the time, it flew at a speed of 740 miles per hour.
What distance did David travel?
2 a. Abigail travels in an airplane a distance of 1860 km. For one-third of the distance, the
airplane flies at a speed of 1032 km/h, and for the rest of the distance, it flies at a speed of
696 km/h. How long does the trip take?
1.
Remember: distance = rate * time
Since he was going at two different rates for each part of the trip, lets first calculate the distance he flew during the first part of the trip and then we can calculate the distance he flew during the second part of the trip. The total distance traveled will be the sum of these two distances.
distance1 = rate1 * time1 = 680 mi/hr * (3.8 / 2) hr = 680 mi/hr * 1.9 hr = 1292 mi
distance2 = rate2 * time2 = 740 mi/hr * (3.8 / 2) hr = 740 mi/hr * 1.9 hr = 1406 mi
total distance = distance1 + distance2 = 1292 mi + 1406 mi = ?
2.
By solving the first formula for time, we get: time = distance / rate
Again let's split this into two parts for each part of the trip.
For the 1st part..... the distance is one third of the total distance, and the rate is 1032 km/hr.
For the 2nd part.... the distance is two thirds of the total distance, and the rate is 696 km/hr.
\(\text{time}_1\ =\ \dfrac{\text{distance}_1}{\text{rate}_1}\ =\ \dfrac{\frac13\cdot1860\ \text{km}}{1032\ \frac{\text{km}}{\text{hr}} } \ =\ \dfrac{620\ \text{km}}{1032\ \frac{\text{km}}{\text{hr}} }\ =\ \dfrac{620}{1032}\ \text{hr}\)
\(\text{time}_2\ =\ \dfrac{\text{distance}_2}{\text{rate}_2}\ =\ \dfrac{\frac23\cdot1860\ \text{km}}{696\ \frac{\text{km}}{\text{hr}} }\ =\ \dfrac{1240\ \text{km}}{696\ \frac{\text{km}}{\text{hr}} }\ =\ \dfrac{1240}{696}\ \text{hr}\)
\(\text{total time}\ =\ \text{time}_1+\text{time}_2 \ =\ \dfrac{620}{1032}\ \text{hr}\ +\ \dfrac{1240}{696}\ \text{hr}\ =\ \Big( \dfrac{620}{1032}\ +\ \dfrac{1240}{696}\Big)\ \text{hr}\ \approx\ 2.38\ \text{hr} \) _