A sprinter reaches his maximum speed in 2.6 seconds from rest with constant acceleration. He then maintains that speed and finishes the 100 yards in the overall time of 10.06 seconds. Determine his maximum speed vmax.

Answer: maximum speed vmax = 11.42m/s

Explanation:

Given that the sprinter maintained constant acceleration during the first 2.6 seconds.

a = vmax/ta ... 1

The distance covered during the acceleration period is;

da = 0.5a (ta) ^2 ... 2

Substituting equation 1 to 2

da = 0.5 (vmax/ta) (ta) ^2 = 0.5vmax (ta) ... 3

The distance covered during the period of constant speed vmax is;

dv = vmax (tv) ... 4

The total distance travelled is

d = da + dv = 100 (Given)

da + dv = 100 ... 5

Substituting equation 3 and 4 into 5

0.5vmax (ta) + vmax (tv) = 100

vmax (0.5ta + tv) = 100

vmax = 100 / (0.5ta + tv) ... 6

But,

t = ta + tv

tv = t - ta ... 7

Substituting equation 6 into equation 7

vmax = 100 / (0.5ta + t - ta)

vmax = 100 / (t-0.5ta)

t = 10.06 s

ta = 2.6 s

Substituting the values;

vmax = 100 / (10.06 - 0.5 (2.6))

vmax = 11.42m/s

Note:

ta = acceleration time

tv = constant velocity vmax time

t = overall time

da, dv and d = acceleration, constant velocity and overall distance covered respectively.