At noon, Ship A is 100 km west of ship B. Ship A travels south at 35 km/h. Ship B travels North at 25 km/h. At 4 pm, how fast the distance between them change?

Answer 1

Make your diagram using an x-y grid
at noon, place ship B at the origin, and A on the left side of the x-axis
After some time of t hrs,
let B's position on the y-axis be Q, mark BQ=25t
draw a vertical line in the South direction and label A's position as P, marking AP=35t

Draw a horizontal line from P to hit the y-axis at R.
Now complete a right-angled triangle QPR, with PR = 100, QR = 25t+35t or 60t

We need d(PQ)/dt

PQ^2 = 100^2 + (60t)^2 = 10000+3600t^2
2 PQ d(PQ)/dt = 7200t

so at 4 pm, t = 4
PQ^@ = 100^2 +240^2
PQ = 260

2(260) d(PQ)/dt = 7200(4)
d(PQ)/dt = 55.38 km/h

check my arithmetic