Question

A sprinter, running a 100-meter race, can accelerate with a constant acceleration for 3.50 s before reaching his top speed. He can then can maintain that speed for the rest of the race. He runs the 100 meters in 10.0 s.

Answer 1

let x = top speed

(3.5 * x/2) + 6.5 x = 100 m ... 8.25 x = 100 m

Answer 2

V = Vo + a*T = 0 + a*3.5 = 3.5a m/s. = velocity reached during acceleration
d1 = 0.5(3.5a*3.5) = 6.125a meters = distance covered during acceleration

d2 = V*T = 3.5a * (10-3.5) = 22.75a meters = dist. covered during last 6.5 s
of race.

d1 + d2 = 100 m.
6.125a + 22.75a = 100
a = 3.5 m/s^2.