1. Feb 15, 2010

### tics

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

A sprinter accelerates from rest to a top speed with an acceleration which magnitude is 3.80 m/s^2 . After achieving top speed, he runs the remainder of the race without speeding up or slowing down. The total race is 50 m long. If the total race is run in 7.88 s, how far does he run during the acceleration phase?

2. Relevant equations

By using equations of motion the following 2 equations arise: s= ut +1/2at^2 = 1/2(3.8)t^2 = 1.9t^2 .......... (1)
From v= u+at and v= s'/t' : s= -3.8t(7.88-t)+50...........(2)

3. The attempt at a solution

The two simultaneous equations lead to :
0= 1.9 ^2-29.94t+50, a quadrqtic equation, which is solved: s= 6.85 m

Can someone please show me step-by-step how they arrived at the final solution. The part that puzzles me the most is how they derived the second equation. Any input is kindly welcomed. Thank you all.

Last edited: Feb 15, 2010
2. Feb 15, 2010

### CompuChip

So the s in the equation, is the distance run in the acceleration time. The first equation approaches it from the acceleration phase: if a = 3.80 m/s/s then equation (1) gives the distance covered during the time t of acceleration.

The second equation looks at the constant velocity part. When you have covered s m of the 50 m race in time t, then the remaining (50 - s) m take (7.88 - t) seconds, at a constant velocity v. This v follows from v = u + a t during the acceleration (with u = 0, a = 3.8 and t again the acceleration time).

From this you can write down a formula in which you isolate s, leading to equation (2). Once you have (1) and (2), you can of course simply equate them and solve for t, then plug back into either of them to find s.

3. Feb 15, 2010

### tics

Thanks a lot CompuChip for your quick reply. And you're right: a scientist is not a person who gives the correct answers, but he's the one who asks the right questions correctly. Keep it up.