Sean is running a 100 m dash. When the starter's pistol fires, he leaves the starting block and continues speeding up until 6 s into the race, when he reaches his top speed of 11 m/s. He holds this speed for 2 s; then his speed has slowed to 10 m/s by the time he crosses the finish line 11 s after he started the race. a. What was Sean's average acceleration during the first 6 s of the race? b. What was Sean's average acceleration from 6 to 8 s into the race? c. What was Sean's average velocity for the whole race? d. What was Sean's average acceleration from 8 to 11 s into the race?

a. a = 1.83 m/s²

b. a = 0 m/s²

c. Average Velocity = 9.09 m/s  

d. a = - 0.33 m/s²

Step-by-step explanation:

a.

We apply 1st equation of motion to the first 6 seconds of the motion:

Vf = Vi + at

a = (Vf - Vi)/t

where,

a = acceleration during first 6 seconds = ?

Vf = Final Speed = 11 m/s

Vi = Initial Speed = 0 m/s

t = time = 6 s

Therefore,

a = (11 m/s - 0 m/s)/6 s

a = 1.83 m/s²

b.

From 6 s to 8 s Sean held his speed constant at 11 m/s. Since, the speed was constant. Therefore, there is no acceleration during this time:

a = 0 m/s²

c.

Sean's average velocity can be found as follows:

Average Velocity = Total Distance Covered/Total Time

Average Velocity = 100 m/11 s

Average Velocity = 9.09 m/s

d.

Applying the 2st equation of motion for last 3 s from 8s to 11 s:

Vf = Vi + at

a = (Vf - Vi)/t

where,

Vf = 10 m/s

Vi = 11 m/s

t = 3 s

Therefore,

a = (10 m/s - 11 m/s)/3 s

a = - 0.33 m/s²

negative sign shows deceleration

answer: y=4000x+14000

step-by-step explanation:

a

step-by-step explanation: