Initial velocity ym
Because it decelerates evenly and stops after 10S, the speed after 10S is 0.
Namely: y- 10x=0.
Get y= 10x.
Then list the equations about y:
10y-0.5× 10? ×x=45
(high school uniform acceleration formula, v0t+gt? /2
T is the time, v0 is the initial velocity, and G is the acceleration (deceleration), that is, X) in the question.
The result is: 10y-50x=45.
Substitute the result just now into the equation about x:
100x-50x=45
x=0.9
Then you get y=9.
So the sliding distance
Precisely
v0t+at? /2=22.5
9t-0.9t? /2=22.5
t? -20t+50=0
Substitute into the root formula:
x= 10-5√2
Another result is giving up.
So the deceleration is
0.9m/s?
Initial speed 45 m/s
After taxiing 10-5√2 seconds, reach the halfway point.
10-5√2
Sorry, the equation is wrong.