# Acceleration of world-class sprinter

1. Aug 18, 2007

### kring_c14

1. The problem statement, all variables and given/known data

A typical world-class sprinter accelerates to his maximum speed in 4 s. If such a runner finishes a 100-m race in 9.1 s, what is the runner’s average acceleration during the first 4 s?

2. Aug 18, 2007

### Pythagorean

this will be a two-part problem.

1) The first part is him accelerating from 0 to Vmax in 4 sec.

2) The second part will be him at steady speed, Vmax for 5.1 sec.

If you can express both of these parts algebraically (without numbers) as meters run, then you can add those together to equal the Total Distance Run and solve for that max velocity.

Having the max velocity, you can go back to the first part and find the acceleration.

3. Aug 18, 2007

### Dick

Divide the race into two parts i) where he accelerates at a constant rate 'a' for 4 sec, and the second part where he travels at the constant speed a*(4 sec). Write down an expression for how far he travels in i) and an expression for how far he travels in ii). The sum of those two expressions is 100m. Can you solve for a?

4. Aug 18, 2007

### kring_c14

i dont get it.....sorry...

5. Aug 18, 2007

### learningphysics

His maximum speed is v which he reaches at 4s... his initial speed is 0. What distance does he travel over the first 4 seconds, in terms of v? Just plug into the appropriate equation.

6. Aug 19, 2007

### Pythagorean

Here's the question right?

For the first part of his motion, he's accelerating from a velocity of 0 to a velocity of V (we don't know the number yet, but that's ok)€.

So we can find the distance run so far by:

1) x1 = 1/2(v1 + v2)t1, where v1 = 0, and t1 = 4s

For the second part, it's simply

2) x2 = v2t2, where we don't know v2 still, but we know t2 = 5.1s

Since the whole race is 100 meters long, and the equations above give us two distances during the time he raced, we can add those two distance, 1) and 2), together to equal the third distance, 100m:

3a) x1 + x2 = 100m

or

3b) 2s*v2 + 5.1s*v2 = 100m

you can solve 3b for v2 to get the velocity.

once you have v2, you can go back to 1) and use the equation:

4) a*t1 = 1/2(v1 + v2)

solving 4) for "a" gives you your acceleration.

7. Aug 19, 2007

### kring_c14

thank you so much!thank you!

8. Aug 19, 2007

### learningphysics

You mean a = (v2-v1)/t1.