having trouble solving these? help!?

Question

A sprinter can accelerate with constant acceleration for 5.06 s before reaching top speed. He can run the 123.0 m in 17.4 s. What is his speed as he crosses the finish line?

and

A car accelerates at 1.95 m/s2 along a straight road. It passes two marks that are 28.0 m apart at times t=3.80 s and t=4.85 s. What was the car's velocity at t=0?

Answer 1

The sprinter's average speed is 7.069 metres per second (123 divided by 17.40). However, he had to work up to that speed. At t=0, his velocity was zero. At t+5.6 seconds, he was at his maximum speed but he could not stay there. For the next 12.34 seconds, he slowed down to a speed of 7.069 metres per second. If his speed is only 7.069 metres per second, for 12.34 seconds, he covers only 80.639 meters, instead of 123 metres. So, he had to go from zero to his top velocity in 5.06 seconds and he had 42.361 metres in which to achieve his top velocity. So, in that 5.06 second burst, he achieved his top velocity of 8.372 metres per second. If he is counting his speed and running at a constant speed, his speed, as he crosses the finish line SHOULD be 7.069 metres per second. But, if he sees his girlfriend taking off her panties as he is about to win, he might speed up a little.

You always start counting from 1, not from zero. So, the velocity of anything at t=0 is always zero. You can't start before you start and t=0 is the starting point. Before anything starts moving, it is not moving at all, meaning that its velocity is zero.

Answer 2

Very interesting problem!
For the accelerating portion:
Vo = 0
∆t = 5.06 sec
Vf² = Vo² + 2a∆D
∆D = (0.5) a(5.06 sec)²
∆D = a (0.5) (25.6036 sec²)
∆D = a (12.8018 sec²)
Now to change the acceleration in the above equation into final velocity:
Vf = Vo + a∆t
Vf = 0 + a (5.06 sec)
a = Vf / 5.06 sec
Substitute it in:
∆D = a (12.8018 sec²)
∆D = (Vf / 5.06 sec) (12.8018 sec²)
∆D = Vf (2.53 sec) is the distance traveled while accelerating.

For the constant speed portion:
∆t = (17.4 sec - 5.06 sec)
∆t = 12.34 sec
∆D = Vo∆T + (0.5) a∆T²
Here, since there is constant speed, the velocity is actually the final velocity of the previous situation, so Vo turns into Vf.
∆D = Vf (12.34 sec)
Add the two equations together:
∆D(1) + ∆D(2) = + Vf (12.34 sec)
We know that ∆D(1) + ∆D(2) = 123 m:
123 m = Vf (2.53 sec) + Vf (12.34 sec)
123 m = Vf (14.87 sec)
Vf = 8.2716879623402824 m/sec is his speed as he crosses the finish line.
Vf = 8.27 m/sec