Back to the topic, my answer is more detailed. You don't need to write so much when you write. Just think about it yourself.

Solution: (1) sδ APD =1/2× ad× AP = 4ap = 4×1× x = 4x As can be seen from Figure (2), when x=a, sδ APD = 4x = 20.

Therefore, x=5, that is, a = 5；; When S δ APD reaches the maximum, point P reaches point B; Point P is from 1 sec to 8 sec, and the passing displacement L=AB-a× 1cm/s=5cm, and the time t=8s-as=8s-5s=3s, so B = L/T = 5/3; When the time is c seconds, the distance from point P to point D is CD+BC= 18cm, so the time from point B to point D is18 ÷ 5/3 = 54/5 =10.8s, so c =10.8+8.

(2) The area variation law of S δ AQD is D-C, increasing, C-B unchanged, and B-A decreasing. From the above, it can be seen that the speed changes to dcm/s at point Q for 5s, and the distance of Q at this time is 5s×2cm/s= 10cm, just reaching point C; According to Figure (3), the time from point C to point A at Q is 22s-5s= 17s, and the distance is CB+BA= 18cm, so the speed is d= 18/ 17.

(3)Y 1=x (x≤5s)

y 1 = 5+(x-5)×5/3(5s & lt； X≤ 18.8 seconds)

Y2=28-2x (x≤5s)

y2 = 18-(x-5)× 18/ 17(5s & lt； x≤22s)

If p and q are satisfied, then Y 1+Y2=AB+BC+CD=28.

When x≤5s, y1= Y2X = 28-2xX = 28/3 > 5. It is impossible to meet;

When in x & gt5s, y1= Y25+(x-5) × 5/3 =18-(x-5) ×18/17, the solution x=9.8 seconds (if the calculation process

(4) 25cm apart, divided into two situations,

1, Y2-Y 1=25 when p and q are not satisfied.

When x≤5s, Y2-Y 1=28-2x-x=25, and the solution is x= 1s, which meets the requirements.

When y2-y1=18-(x-5) ×18/17-5-(x-5 )× 5/3 = 25 in x & gt5s, the solution does not meet the requirements;

2. When p and q meet, Y 1-Y2=25.

When x≤5s, Y 1-Y2=x-28+2x=25, and the solution is x= 17, which does not meet the requirements.

When y1-y2 = 5+(x-5) × 5/3-18+(x-5) ×18/17 = 25 in x & gt5s, the solution x =/kloc-0.

So at 1s, the distance is 25cm.

Finally, I want to spit out the data used in the calculation.