# Velocity and Acceleration

1. Sep 1, 2005

### Tabe

Ok, I have this problem...
The first 10 meters of a 100 meter dash are covered in 2 seconds by a sprinter who starts from rest and acceleraties with a constant acceleration. The remaining 90 meters are run with the same velocity the stprinter had after 2 seconds.
a) Determin the sprinter's constant acceleration during the first 2 seconds.
b) Determine the sprinter's velocity after 2 seconds have elapsed.
c) Determine the total time needed to run the full 100 meters.

What I don't understand is the 'accelerates with a constant acceleration' part. All I did to solve that part was solve for the average acceleration, which I think is wrong.
I ended up getting 2.5 m/s^2 for the acceleration.

Could anyone tell me whether or not I'm on the right track, or if I'm totally wrong, just point me in the right direction.

2. Sep 1, 2005

### quasar987

Average acceleration? How did you get that? Anyway, you know the acceleration of the sprinter is constant, and you know the equation of position wrt time for an object at constant acceleration:

$$x(t) = x_0 + v_0t + 0.5at^2$$

Use that and the data you know to retrieve a.

P.S. You should have posted this in the Homework section: https://www.physicsforums.com/forumdisplay.php?f=15

Even if it's not officially an homework, it's clearly a question taken from a textbook.

Last edited: Sep 1, 2005
3. Sep 1, 2005

### Tabe

Ok, I used that formula, and ended up getting 5 m/s^2. Is that what I should have gotten or did I mess up again somewhere.

4. Sep 1, 2005

### quasar987

I get that too.

5. Sep 1, 2005

### Tabe

So, if I use 5m/s^2 and solve for velocity, I get 10m/s, then when I solve for the time, I get 10s. Is that right, because I got really confused when it mentioned constant acceleration and velocity.

6. Sep 1, 2005

### quasar987

Yes for speed, nope for time. Traveling 90 meters at speed 10m/s takes 9 seconds. And before that, he was accelerating for 2 seconds. So total time to complete the race: 9 + 2 = 11s.

What confuses you about the notions of constant acceleration and constant velocity? Perhaps we can sort that out.

7. Sep 1, 2005

### Mk

I accel. Get it? excel = accel. Ha ha I'm so lame.

8. Sep 1, 2005

### Tabe

I really mix up and/or combine velocity and acceleration. The change from constant acceleration to constant velocity really confused me, because I know that acceleration can remain constant.

9. Sep 1, 2005

### quasar987

I hear ya. Acceleration is the building of velocity, as told by the units of acceleration: x m/s² = x (m/s)/s. I.e. the object's speed gains x m/s every second. So if the acceleration is constant, the velocity is building at a constant rate. That is how the two are related. On the other hand, if the velocity is constant, it necessarily means that the acceleration is 0 (no building of velocity).