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
