Sprinter Speed at Finish Line in 100 Meter Dash

[TonkaQD4]

A sprinter can accelerate with constant acceleration for 4.0s before reaching top speed. He can run the 100-meter dash in 10s. What is his speed as he crosses the finish line?

I need help with this problem.

I know that for the last 6 seconds because he has reached top speed that he has constant velocity. During the first 4 seconds I must use a Uniformed Accelerated Motion formula, and I think I need to actually combine 2 of them together; however, I get an unrealistic answer, like 15m/s, which I know is incorrect.

Please help me find the sprinter's speed as he crosses the finish line?

 

[Avodyne]

Show us how you got 15m/s. (Which is not crazy, by the way. Since 100m is convered in 10s, the average speed is 10m/s. The speed at the end must be greater than this, since it starts off at zero.)

 

[TonkaQD4]

Actually I change my answer... 12.5m/s

But I am still a little skeptical about one step, I actually used another person's related forum to solve some of it and at the part I get stuck I used his/her but I am not quite sure if it is right.

First
x= xi + vi(delta t) + .5 a(delta t^2) UAM FORMULA
x= 0+0+.5a(4^2)
x= 8a

Then
Vf= Vi + a(delta t)
Vf= 0 + a(4s)

Therefore 4a= (100-x)6

And

4a= (100-x)6
4a= (100-8a)6 <--this is where I used another source and get confused on the next step
32a = 100 ? is this correct??
a= 3.125


Now Final V

Vf= 4s times 3.125m/s

Vf = 12.5m/s

The sprinters speed as he crosses the finish line is 12.5m/s.

Please Help! Is this the correct answer or approach to this problem.

 

[Avodyne]

Almost. You've got the 6 on the wrong side in 4a=(100-x)6; it should be (4a)6=100-x, because the velocity is 4a, the time is 6, and you multiply these to get the distance, which is 100-x. With this correct form, you then have 24a=100-x=100-8a, or 32a=100.
And 12.5m/s is the correct final answer.