A sprinter begins from rest and accelerates at the rate of 2 m/s^2 for 200 m.
a. What is the sprinters velocity at the end of the 200 m?
b. How long does it take him to cover it?
c. What is his average velocity?

Answers

Answer 1

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

Answer 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.

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