The displacement of an object is S1 in the first 3 seconds and S2 in the last 3 seconds The displacement of an object in the first 3 seconds is S1, and that in the last 3 seconds is S2. Given s2-s1 = 6m, S1: S2 = 3:7, the total length of the inclined plane can be calculated The slide is accelerating. Let the length of slope be l and the total time be t L=a*t^2 / 2 S1=a *3^2 / 2 S2=L-[ a*( t -3)^2 / 2 ]=3at-4.5a 3/7=(a *3^2 / 2)/(3at-4.5a) T = 5 seconds 6=(3at-4.5a)-(a *3^2 / 2) a=1m/s^2 L = 12.5A = 12.5m

The displacement of an object is S1 in the first 3 seconds and S2 in the last 3 seconds The displacement of an object in the first 3 seconds is S1, and that in the last 3 seconds is S2. Given s2-s1 = 6m, S1: S2 = 3:7, the total length of the inclined plane can be calculated The slide is accelerating. Let the length of slope be l and the total time be t L=a*t^2 / 2 S1=a *3^2 / 2 S2=L-[ a*( t -3)^2 / 2 ]=3at-4.5a 3/7=(a *3^2 / 2)/(3at-4.5a) T = 5 seconds 6=(3at-4.5a)-(a *3^2 / 2) a=1m/s^2 L = 12.5A = 12.5m

12.5, it's really a bit of trouble, I don't know if it's right
Key uniform acceleration motion, initially and finally for 3 seconds, with the image solution convenient