a sprinter accelerates from rest to a velocity of 12m/s in the first 6 seconds of the 100 meter dash .
a ) how far does the sprinter travel during the first 6 seconds ?
b ) how much farther does the sprinter have to travel to reach the finish line ?

Respuesta :

Answer:

a) 36 m

b) 64 m

Explanation:

Given:

v₀ = 0 m/2

v = 12 m/s

t = 6 s

Find: Δx

Δx = ½ (v + v₀) t

Δx = ½ (12 m/s + 0 m/s) (6 s)

Δx = 36 m

The track is 100 m, so the sprinter still has to run another 64 m.

If a sprinter accelerates from rest to a velocity of 12m/s in the first 6 seconds of the 100-meter dash, then the sprinter travels 72 m  during the first 6 seconds and the sprinter has to travel 28 m more to reach the finish line

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

Keep in mind that these calculations only apply to uniform acceleration.

As given in the problem,a sprinter accelerates from rest to a velocity of 12m/s in the first 6 seconds of the 100-meter dash

By using the first equation of the motion

v = u + at

12 = 0+a*6

a = 2 m/s²

initial velocity(u) = 0 m/s

acceleration(a) = 9.81 m/s²

By using the second equation of motion

S = ut + 1/2at²

,u= 0 m/s , a= 2m/s² and t =6 seconds

S = 0 +2×6²

S =72 meter

As he has to cover a total of 100 meters, the sprinter has to travel 28 m more to reach the finish line

Thus, the sprinter travels 72 m  during the first 6 seconds and the sprinter has to travel 28 m more to reach the finish line

Learn more about equations of motion from here

brainly.com/question/5955789

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