Ship A leaves Eastport Harbor at 1018 am on a course given b

Ship A leaves Eastport Harbor at 10:18 am on a course given by the equation y = (4/6)x + 3. Eastport Harbor is located at coordinates (-3, 1). Earlier at 10:02 am Ship B left Westport Harbor on a course given by the equation y = (-7/5)x + 46/5. Westport Harbor is located at coordinates (8, -2).

The speed of Ship A is 12 mph, and the speed of Ship B is 10 mph. All distances are in miles.

Is it likely that these two ships will collide?

If a collision is likely what would be the time and location of the collision?

Solution

Now ship A course --- y = (4/6)x + 3

ship B course ---- y = (-7/5)x + 46/5

Ship would collide at the point of intersection of these two lines:

(4/6)x + 3 = (-7/5)x + 46/5

x{ 4/6 + 7/5 } = 46/5 -3

x(20+42)/30 = ( 46 -15)/5

x( 62/30) = 31/5

x = 31*30/(5*62) = 3

y = (4/6)*3 + 3 = 2 +3 =5

Location = ( 3,5)

Now we find the distance traveled by ship A between ( -3, 1) to ( 3, 5) = sqrt{( 5-1)^2 +(3+3)^2}

=sqrt( 16 + 36) = sqrt52 miles

Time taken = distance/speed = sqrt(52)/12 = 0.6 hrs = 36 minutes later from starting time of ship A 10: 18 am

Time = 10 : (18+36) = 10 : 54 am