USE DISTANECE= RATE * TIME

1- Anjali traveled to teh recycling plant and back. The trip there took 5.8 hrs and the trip back took 5.1 hrs. She averaged 7 mph faster on the return trip than the outbound trip. What was Anjali's average speed on the outbound trip?

2- A fishing boat left Hawaii traveling west 0.5 hrs before a cruise ship. the cruise ship traveled in the opposite direction going 12.5km/h faster than the fishing boat for 11 hrs after which the time the ships were 322 km apart. What was te fishing boat's speed.

a]
D=R×t
d=5.8r
d2=5.1(r+7)
0.7r=35.7
r=51 mi/hr
thus the average speed on the outbound trip would be:
51+7=58 mi/hr

b]
let speed of fishing boat=x km/h
speed of cruise ship=(x+12.5) km/h
distance traveled by fishing boat=11.5×x=11.5x km
distance traveled by cruise ship=(x+12.5)×11=(11x+137.5) km
Total distance covered by the ships:
11.5x+11x+137.5=322
22.5x=322-137.5
22.5x=184.5
thus
x=184.5/22.5
x=8.2 km/h
The speed of fishing boat is 8.2 km/hr

eudora eudora · Answer 2 · 2025-03-17T20:40:22-04:00

Answer:

1). Anjali's speed on the outbound trip was 5.1 miles per hour.

2). Speed of the boat was 8.2 km per hour.

Step-by-step explanation:

1). Anjali traveled to teh recycling plant in 5.8 hrs and trip back took 5.1 hrs.

Her speed was 7 mph faster on the return trip.

Let her outbound trip speed was = s mph

Then her speed for the return trip will be = (s + 7) mph

Distance traveled to plant = Speed × time

= s×5.8 miles

Distance traveled in the back trip = (s + 7)×5.1 miles

Since distances covered to outbound trip and return trip are same

So, 5.8s = (s + 7)×5.1

5.8 s = 5.1 s + 7×5.1

5.8 s - 5.1 s = 35.7

0.7 s = 35.7

s = [tex]\frac{35.7}{0.7}[/tex]

= 5.1 miles per hour

Therefore, Anjali's speed on the outbound trip was 5.1 miles per hour.

2). Let the speed of the ship is s km per hour

Ship traveled for 11 hours so distance covered by the ship = Speed × time = s×11

Since speed of the ship was 12.5 km per hour faster than the boat then

Speed of the boat will be = s - 12.5

Since boat started traveling 0.5 hrs before the ship then total time the boat traveled = 11 + 0.5 = 11.5 hours

Now the distance covered by the boat = (s - 12.5)×11.5

Since distance between the ship and the boat is 322 km so the equation will be

11s + (s - 12.5)11.5 = 322

11s + 11.5s - 12.5×11.5 = 322

22.5s - 143.75 = 322

22.5s = 322 + 143.75

22.5s = 465.75

s = [tex]\frac{465.75}{22.5}[/tex]

= 20.7 km per hour

Now the speed of the boat will be = (s - 12.5)

= (20.7 - 12.5)

= 8.2 km per hour