Respuesta :
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.
