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.