at 12 noon ship A is 65 km due north of a second ship B. Ship A sails south at a rate of 14km/hr, and ship B sails west at a rate of 16km/hr. How fast are the two ships approaching each other 1.5 hours later at 1:30pm? Thank You!
To find out how fast the two ships are approaching each other, we need to calculate their relative velocity.
Let's start by finding the positions of the ships at 1:30 pm, which is 1.5 hours after 12 noon.
- Ship A traveled south at a rate of 14 km/hr for 1.5 hours, so it covered a distance of 14 km/hr * 1.5 hr = 21 km.
- Ship B traveled west at a rate of 16 km/hr for 1.5 hours, so it covered a distance of 16 km/hr * 1.5 hr = 24 km.
Now, we can calculate the positions of the two ships at 1:30 pm:
- Ship A is 65 km (initial distance) - 21 km (distance traveled south) = 44 km north of Ship B.
- Ship B is still in the same position since it only traveled west.
We can now find the distance between the two ships at 1:30 pm. Since Ship A is north of Ship B, we can consider their positions as forming a right-angled triangle. By using the Pythagorean theorem, we have:
Distance^2 = (44 km)^2 + (24 km)^2
Distance^2 = 1936 km^2 + 576 km^2
Distance^2 = 2512 km^2
Distance ≈ 50.12 km
Now, we know the distance between the two ships is approximately 50.12 km. Since the time elapsed was 1.5 hours, we can find the relative velocity of the ships by dividing the distance by the time:
Relative Velocity = 50.12 km / 1.5 hr ≈ 33.42 km/hr
Therefore, the two ships are approaching each other at a speed of approximately 33.42 km/hr.
Making a sketch for the 12 noon position, placing B at the origin and A 65 units up on the y-axis
At a time of t hrs after noon,
let the distance covered by A be 14t
let the distance covered by B be 16t
Draw a line between their positions and call it d km
d^2 = (16t)^2 + (65-14t)^2
when t = 1.5
d^2 = 620
d = √620
2d dd/dt = 2(16t)(16) + 2(65- 14t)(14)
dd/dt = (256t + 65 - 196t)/√620
when t=1.5
dd/dt = (256(1.5) + 65 - 196(1.5))/√620
= 6.22 km/h
check my arithmetic