SOLUTION: Imagine a sprinter accelerates from rest to maximum speed of 10.8 m/s in 1.8s.in what time interval will he finish. 100 meter race if he keeps the speed constant at 10. 8 m/s for t

Question 896852: Imagine a sprinter accelerates from rest to maximum speed of 10.8 m/s in 1.8s.in what time interval will he finish. 100 meter race if he keeps the speed constant at 10. 8 m/s for the last part of the race what assumpitions did you make? Please show your work.

Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
the formula for distance under acceleration is:

d = v*t + 1/2 * a * t^2

d = distance traveled
v = initial velocity
t = time
a = acceleration.

you also need the acceleration formula of:

a = change in velocity divided by change in time.

the runner accelerates from 0 velocity to 10.8 meters per second in 1.8 seconds.

the acceleration formula is a = change in velocity divided by time.

change in velocity is from 0 to 10.8 meters per second which is equal to 10.8

the time is 1.8 seconds.

the acceleration is 10.8/1.8 = 6 meters per second squared.

now we can find the distance traveled during the acceleration.

the formula is d = v*t + 1/2 * a * t^2

v*t = 0 because the initial velocity is equal to 0.

the formula becomes:

d = 0 + 1/2 * a * t^2

a is equal to 6 meters per second squared.
t^2 is equal to 1.8^2.

formula becomes:

d = 0 + 1/2 * 6 * 1.8^2 which is equal to 9.72 meters.

the runner has run 9.72 meters during the runner's acceleration phase.

since the race is 110 meters, the runner has another 100.28 meters to go.

since rate * time = distance, the formula becomes:

10.8 meters per second * time = 100.28 meters.

solve for time to get time = 100.28 / 10.8 = 9.29 seconds rounded to the nearest one hundredth of a second.

add that to the 1.8 second that the runner took to get up to speed and the total time the runner took to finish the race is is 9.29 + 1.8 = 11.09 seconds.

that's your solution if i did this correctly.

total time is 11.09 seconds.

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