Question

1. ryan left the science museum and drove south. gabriella left three hours later driving 12 km/h faster in an effort to catch up to him. after two hours gabriella finally caught up. find ryan’s average speed.
2. a submarine left hawaii two hours before an aircraft carrier. the vessels traveled in opposite directions. the aircraft carrier traveled at 25 mph for nine hours. after this time the vessels were 280 mi. apart. find the submarine’s speed.
3. chelsea left the white house and traveled toward the capital at an average speed of 34 km/h. jasmine left at the same time and traveled in the opposite direction with an average speed of 65 km/h. find the number of hours jasmine needs to travel before they are 59.4 km apart.
4. jose left the airport and traveled toward the mountains. kayla left 2.1 hours later traveling 35 mph faster in an effort to catch up to him. after 1.2 hours kayla finally caught up. find jose’s average speed.

Explanation:

Problem 7:

Step1: Define Variables

Let Ryan's speed be \( x \) km/h. Then Gabriella's speed is \( x + 12 \) km/h.
Ryan's time of travel: \( 3 + 2 = 5 \) hours (since he left 3 hours earlier and Gabriella caught up after 2 hours).
Gabriella's time of travel: \( 2 \) hours.

Step2: Set Up Distance Equation

Distance = Speed × Time. Since they traveled the same distance when Gabriella caught up:
\( 5x = 2(x + 12) \)

Step3: Solve the Equation

Expand: \( 5x = 2x + 24 \)
Subtract \( 2x \): \( 3x = 24 \)
Divide: \( x = 8 \)

Step1: Define Variables

Let the submarine's speed be \( x \) mph. Then the aircraft carrier's speed is \( x + 25 \) mph (since it's faster? Wait, no—opposite directions, so relative speed? Wait, no: submarine left 2 hours earlier, then aircraft carrier left. Wait, time: submarine's time: \( 2 + 9 = 11 \) hours. Aircraft carrier's time: \( 9 \) hours. Distance apart: 280 mi. Wait, opposite directions, so total distance is sum of their distances.
Wait, re - read: "A submarine left Hawaii two hours before an aircraft carrier. The vessels traveled in opposite directions. The aircraft carrier traveled at 25 mph for nine hours. After this time the vessels were 280 mi. apart. Find the submarine’s speed."

So submarine's time: \( 2 + 9 = 11 \) hours. Speed: \( x \) mph. Distance: \( 11x \)
Aircraft carrier's time: 9 hours. Speed: 25 mph. Distance: \( 25×9 = 225 \) mi.
Total distance: \( 11x + 225 = 280 \)

Step2: Solve for \( x \)

\( 11x = 280 - 225 = 55 \)
\( x = \frac{55}{11}=5 \)

Step1: Define Variables

Let \( t \) be the time (in hours) they traveled until they were 59.4 km apart.
Chelsea's speed: 34 km/h, distance: \( 34t \)
Jasmine's speed: 65 km/h, distance: \( 65t \)
Since opposite directions, total distance: \( 34t + 65t = 59.4 \)

Step2: Solve for \( t \)

\( 99t = 59.4 \)
\( t=\frac{59.4}{99}=0.6 \) hours. Wait, but the question says "Find the number of hours Jasmine needs to travel before they are 59.4 km apart." Wait, Chelsea left, Jasmine left at same time? Wait, re - read: "Chelsea left the White House and traveled toward the capital at an average speed of 34 km/h. Jasmine left at the same time and traveled in the opposite direction with an average speed of 65 km/h. Find the number of hours Jasmine needs to travel before they are 59.4 km apart."

So same time \( t \). So \( 34t + 65t = 59.4 \), \( 99t = 59.4 \), \( t = 0.6 \) hours.

Answer:

Ryan's average speed is 8 km/h.

Problem 8: