100 m dash runner starts at rest %26amp; accelerates at constant rate for 40 meters before having acceleration of 0 for 60 meters; total time is 9.85 s --- THANKS!
What is the a) top speed, b) acceleration for first part, c) time per part?
Lets call the distance of the first part S1 and the total distance S2, and the corresponding times T1 and T2. We know:
S1 = 40
S2 = 100
T2 = 9.85
If V is the maximum velocity we know S1 = 0.5VT1 (const acc.)
For the second part we know S2-S1 = V(T2-T1) = VT2 - VT1
VT1 = 2S1 so S2-S1 = VT2 - 2S1 or V = (S2+S1)/T2 = 140/9.85
So a) - V = 14.21 m/s
We know V^2 = 2AS1 where A is acceleration so A = V^2./2S1
or A = 14.21^2 /(2x40)
So b) A = 2.525 m/s^2
We know V = AT1 so T1 = V/A = 14.21/2.525 = 5.62 s
So c) the two parts are 5.62s and 4.23s
survey
Subscribe to:
Post Comments (Atom)
No comments:
Post a Comment