Question
A runner travels 100 m in 10.9 s. Assume that it takes him 1.35 s to reach his top speed, which is then maintained for the rest of the race. Find the following values:
a) acceleration during the first 1.35 s
b) maximum speed
a) acceleration during the first 1.35 s
b) maximum speed
Answers
drwls
(1/2)*a*(1.35)^2 + Vmax*(9.55) = 100 m
You need a second equation to solve for two unknowns. Here it is:
a*1.35 = Vmax
Therefore
(1/2)*a*(1.35)^2 + 12.89 a = 100
13.80 a = 100
a = 7.246 m/s^2
Vmax = 9.78 m/s
You need a second equation to solve for two unknowns. Here it is:
a*1.35 = Vmax
Therefore
(1/2)*a*(1.35)^2 + 12.89 a = 100
13.80 a = 100
a = 7.246 m/s^2
Vmax = 9.78 m/s