Answer:
a. 28.28
B.14.4
C.14.4
Explanation:
A. v^2=u^ + 2as
v^2=0^2 + 2*2*200
B. v=u+at
t=v-u/a
C. V+u/2
The sprinter velocity at the end of the 200 m would be 28.28 m/s.
What are the three equations of motion?There are three equations of motion given by Newton
The first equation is given as follows
v = u + at
the second equation is given as follows
S = ut + 1/2×a×t²
the third equation is given as follows
v² - u² = 2×a×s
As given in the problem a sprinter begins from rest and accelerates at the rate of 2 m/s^2 for 200 m.
By using the third equation of the motion,
v² - u² = 2×a×s
v² - 0² = 2×2×200
v² = 800
v =28.28 m/s
The sprinter velocity at the end of the 200 m would be 28.28 m/s
By using the second equation of motion
S = ut + 1/2*a*t²
u= 0 m/s , a= m/s² s = 200 m
200 = 0 + 0.5*2*t²
t² = 200
t = √200
t = 14.14 seconds
It would take him 14.14 seconds to cover.
average velocity = initial velocity + final velocity /2
= 0+ 28.28/2 m/s
= 14.14 m/s
Thus, The sprinter velocity at the end of the 200 m would be 28.28 m/s.
Learn more about equations of motion from here
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