# Need some guidance through a dynamics problem.

## 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 travelled 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.

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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 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?

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 travelled = 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?

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 travelled = 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.