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A world-class sprinter accelerates to his maximum speed in 4.0s. He then maintains this speed for the remainder of a 100m race, finishing with a total time of 9.1s. (a) What is the runner's average acceleration during the first 4.0s? (b) What is is average acceleration during the last 5.1s? (c) What is his average acceleration for the entire race? (d) Explain why your answer to part (c) is not the average of the answers to parts (a) and (b)

Added Mon, 06 Jul '15

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There are two phases in this problem, one is when the runner reaches his max. speed at 4 secs and when his max. speed becomes the constant speed for the last 5.1 secs, thus having a total time of 9.1 secs running a 100-m race.

We all know that the distance of phase 1 plus the distance of phase 2 is equal to the total distance, which is 100 m.

s1 + s2 = s(total)

s1 + s2 = 100

Phase 1:

Vi1 = 0

t1 = 4 s

a1 = ?

Vf1 = ?

Let's solve for s1:

s1 = (Vi1 + Vf1)/2 * t1

s1 = (0 + Vf1)/2 * 4

s1 = 2Vf1

Phase 2:

v = Vf1

t2 = 5.1 s

s2 = Vf1*t2

s2 = 5.1Vf1

s1 + s2 = s(total)

2Vf1 + 5.1Vf1 = 100

7Vf1 = 100

Vf1 = 14.085 m/s

After determining Vf1, we can now solve for the acceleration at:

a.) First 4 secs:

14.085 = 0 + a(4)

a = 3.521 m/s^2

b.) Last 5.1 secs

a = 0 [since he maintains his velocity for the remainder of the race.]

<< What is his average acceleration for the entire race? >>

Average acceleration = (3.521*4 + 0)/9.1 = 1.547 m/sec.

<< Explain why your answer to part c is not the average of the answers to parts (a) and (b)? >>

Since there are two distinct times involved (4 sec and 5.1 sec), these have to be taken into consideration in determining tha average acceleration.

The only time the answer to this part would be equal to the average of the answers in Parts (a) and (b) is when the times for the phases of the problem are the same, i.e., time for runner to reach maximum speed = time for runner to finish the rest of the race distance.

Added Mon, 06 Jul '15

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