Consult Concept Simulation 21 before starting this problem T

Consult Concept Simulation 2.1 before starting this problem. The Kentucky Derby is held at the Churchill Downs track in Louisville, Kentucky. The track is one and one-quarter miles in length. One of the most famous horses to win this event was Secretariat. In 1973 he set a Derby record that has never been broken. His average acceleration during the last four quarter-miles of the race was +0.0105 m/s². His velocity at the start of the final mile was about +16.58 m/s. The acceleration, although small, was very important to his victory. To assess its effect, determine the difference between the time he would have taken to run the final mile at a constant velocity of +16.58 m/s and the time he actually took. Although the track is oval in shape, assume it is straight for the purpose of this problem.
I need this to be in the correct significant figures and units as well.

Solution

here in the question distance is not given so i take 1609 m or you can take any alphabet or the value given in the question .. for any value of distance the same formulas will be use

Let T1 = time it will take Secretariat to travel 1609 m at constant velocity of 16.58 m/s.

T1 = D/V = 1609 /16.58

T1 = 97.0sec

Let T2 = time it will take Secretariat to travel 1609 m if his average acceleration is 0.0105 m/s^2.

D = Vi * (T2) + (1/2) * (a) *(T2)^2

1609 = (16.58) * (T2) + (1/2) * (0.0105) * (T2)^2

1609 = 16.58 * (T2) + 0.00525 * (T2)^2

0.00525 * (T2)^2 + 16.58 * (T2) - 1609 = 0

Using the Quadratic Formula to solve for T2, we have

T2={- 16.58 +/- sqrt[(16.58)^2 - 4(0.00525) (-1609)] / [2(0.00525)]}

T2 = 94.2 sec

Time Difference = T1 - T2 = 97.0 - 94.2 = 2.8 sec

the difference between the time he would have taken to run the final mile at a constant velocity of +16.58 m/s and the time he actually took is 2.8sec