1. Let the coordinates of n be N 1 and N2AC = 5 respectively. Because m reaches the destination first, in the triangle AOC, the included angle of sin OAC=OC/AC=(OC-N2)/CN is 4/5=(4-N2)/t, and the Y-axis coordinate N2 is 4-4t/t.

That is N 1=3t/5, that is, the n coordinate is (3t/5,4-4t/5) 0.

2. The area is equal to the sum of triangle AON and AMN S =1/2 (OA * N2+AM * (OA-N1) =1/2 * [3 * (4-4t/5)+t * (3-3t/5).

That is, s =-0.3t2+0.3t2+6.

3.tan angle AOM=N2/N 1=AM/OA, that is, (4-4t/5)/(3t/5)=AM/AO, the simplified solution is T 2+4t-20 = 0, and the solution is the square root of 6-2 of t=2.

It's hard to explain here, so I'll just say it briefly.