Two boat leave a port at the same time the first travels at 15km /hr on a bearing 135 while the second travel at 20km /hr on a bearing North of the first boat calculate their distance a part
Please i need the answer now please i beg u in the name of God
This whole thing is garbled.
you want distances, but give only speeds. distance = speed * time.
ships travel on headings, not bearings.
and what the heck does "on a bearing North of the first boat" even mean?
I could probably figure it all out, but it's not my job to figure out what you mean. It's your job to say it so clearly you cannot be misunderstood. Especially in math, where accuracy is so critical.
answer
To calculate the distance between the two boats, we can use the concept of relative velocity.
First, let's find the components of motion for both boats:
Boat 1:
- Speed: 15 km/hr
- Bearing: 135 degrees
Boat 2:
- Speed: 20 km/hr
- Bearing: North (which is 0 degrees)
To find the components of motion for Boat 1:
- The horizontal component = speed * cos(angle)
= 15 km/hr * cos(135 degrees)
≈ -10.61 km/hr (negative because it's going to the left)
- The vertical component = speed * sin(angle)
= 15 km/hr * sin(135 degrees)
≈ 10.61 km/hr
To find the components of motion for Boat 2:
- The horizontal component = speed * cos(angle)
= 20 km/hr * cos(0 degrees)
= 20 km/hr
- The vertical component = speed * sin(angle)
= 20 km/hr * sin(0 degrees)
= 0 km/hr
Now, let's calculate the distance between the two boats using the Pythagorean theorem:
Distance = sqrt((horizontal component2 - horizontal component1)^2 + (vertical component2 - vertical component1)^2)
Plugging in the values:
Distance = sqrt((20 km/hr - (-10.61 km/hr))^2 + (0 km/hr - 10.61 km/hr)^2)
Distance = sqrt((30.61 km/hr)^2 + (-10.61 km/hr)^2)
Distance ≈ sqrt(937.0921 km^2 + 112.5721 km^2)
Distance ≈ sqrt(1049.6642 km^2)
Distance ≈ 32.40 km
Therefore, the two boats are approximately 32.40 km apart.