Question

ou are writing a short adventure story for your English class. In your story, two submarines need to arrive at a place in the middle of the Atlantic Ocean at the same time. They start out at the same time from positions equally distant from the rendezvouspoint. They travel at different speeds, but both go in a straight line. The first submarine travels at an average speed of 20 km/hr for the first 500 km, 40 km/hr for the next 500 km, 30 km/hr for the next 500 km and 50 km/hr for the final 500 km. In the story’s plot, the second submarine is required to travel at a constant velocity, so the captain needs to determine the magnitude of that velocity. What is that velocity?

Answer 1

The second submarine must maintain a constant velocity of 31.17 km/h to arrive at the rendezvous point at the same time as the first submarine, which took a total of 64.17 hours to travel 2000 km at varying speeds.

Step-by-step explanation:

To determine the constant velocity the second submarine must travel at to arrive at the rendezvous point in the Atlantic Ocean at the same time as the first submarine, we must first calculate the total time taken by the first submarine. This involves calculating the time for each leg of its journey and then summing these times.

The first submarine travels four legs at different speeds:

1. 500 km at 20 km/h, which takes 25 hours,
2. 500 km at 40 km/h, which takes 12.5 hours,
3. 500 km at 30 km/h, which takes 16.67 hours,
4. 500 km at 50 km/h, which takes 10 hours.

The total time taken for the journey is therefore 25 + 12.5 + 16.67 + 10 = 64.17 hours. The total distance traveled by the first submarine is 2000 km (500 km for each leg).

To arrive at the same time, the second submarine must also take 64.17 hours to travel the same distance. The required constant velocity of the second submarine is the total distance divided by the total time:

Velocity = Total distance / Total time = 2000 km / 64.17 hours = 31.17 km/h.

The second submarine must travel at a constant velocity of 31.17 km/h to arrive concurrently with the first.

Answer 2

The constant speed of second submarine is 31.16 km/hr

Step-by-step explanation:

Given that

v₁=20 km/hr ,d₁= 500 Km

v₂=40 km/hr ,d₂=500 km

v₃=30 km/hr, d₃=500 km

v₄=50 km/hr ,d₄=500 km

We know that

Displacement = Velocity x Time

d = v t

Total displacement = Average velocity x Total time

`d_1+d_2+d_3+d_4=V_(avg)\left((d_1)/(v_1)+(d_2)/(v_2)+(d_3)/(v_3)+(d_4)/(v_4)\right)`

Now by putting the values

`2000=V_(avg)\left((500)/(20)+(500)/(40)+(500)/(30)+(500)/(50)\right)`

`V_(avg)=31.16\ km/hr`

So the constant speed of second submarine will be the average speed of first submarine because they have to meet at the same time.

The constant speed of second submarine is 31.16 km/hr