Need some guidance through a dynamics problem.

[vvvether]

Homework Statement

A 50kg sprinter ran a 100m dash in 10.97s. This sprinter set the record for the 100m dash at normal altitudes. Assume that she accelerated uniformly for the first 12.0m and then traveled at a constant velocity to the finish line.

Homework Equations

a) What is the sprinter's maximum velocity
b) What average horizontal force did the sprinter's feet exert on the ground while accelerating?

The Attempt at a Solution

I don't really know where to start tackling this problem.

EDIT: Problem solved. I only really had trouble with finding the terminal velocity. The second part wasn't too hard.

 

[truesearch]

I like to see this shown as a velocity~time graph. In the first section the velocity increases uniformely up to v in t secs. In the second section the velocity remains at v for a time of (10.97-t) secs.
The total distance (100m) is made up of the 2 distances in section 1 and section 2.
Can you see how to get an equation to get v?
(I got 10.21 m/s)

 

[vvvether]

I like to see this shown as a velocity~time graph. In the first section the velocity increases uniformely up to v in t secs. In the second section the velocity remains at v for a time of (10.97-t) secs.
The total distance (100m) is made up of the 2 distances in section 1 and section 2.
Can you see how to get an equation to get v?
(I got 10.21 m/s)

I don't, unfortunately... I've tried the problem for quite some time now. Could you maybe go into more detail?

 

[truesearch]

Take the first section...constant acceleration from 0 to v m/s. so the average velocity for this part is v/2. If we say it takes t seconds then the distance traveled = t x v/2.
And we know this distance is 12m so (vt)/2 = 12 or vt = 24
For the second part the velocity (v) is constant and the time taken is (10.97-t) and the distance is (100-12) =88m.
Can you have a go at writing the equation for the second part then use the vt = 24 to solve it?

 

[vvvether]

Take the first section...constant acceleration from 0 to v m/s. so the average velocity for this part is v/2. If we say it takes t seconds then the distance traveled = t x v/2.
And we know this distance is 12m so (vt)/2 = 12 or vt = 24
For the second part the velocity (v) is constant and the time taken is (10.97-t) and the distance is (100-12) =88m.
Can you have a go at writing the equation for the second part then use the vt = 24 to solve it?

Okay, the second equation is v2 = 88/(10.97 - t1), which I had no trouble getting to. The fact that you helped make the first equation made it click for me.

I might as well solve the problem here, if other people need it.

Since vt = 24, we can isolate for v. Which means v = 24/t. Then we take the second equation, v = 88/(10.97 - t), and sub v = 24/t in place of v.

So 24/t = 88/(10.97 - t). Then I solve for t, which I do by cross multiplying:

24(10.97 - t) = 88t
261.28 - 24t = 88t
261.28 = 88t + 24t
261.28 = 112t
t = 2.35

Sub in t into the either equation to get the terminal velocity.

v = 24/t
v = 24/2.35
v = 10.21 m/s

EDIT: Thanks a lot by the way.