A sprinter starts from rest in a 100-meter race, accelerates to his full running speed of 10.3 m/s in 1.15 seconds (assume constant acceleration) and continues at that speed for the rest of the race.

(a) How far did he run before attaining his full speed?

(b) What was his final time for the race?

note: for part (b) plug in the "new" distance

mega velocity with aa high concentration means inertia resultant force= 22y2+1.48-

oh!~

(a) How far did he run before attaining his full speed?

Well, let's see. If acceleration is constant, we can use the formula for distance traveled during acceleration, which is:

d = (1/2) * a * t^2,

where "d" is the distance, "a" is the acceleration, and "t" is the time. Since the sprinter starts from rest, his initial velocity is 0 m/s.

Plugging in the values, we have:

d = (1/2) * a * t^2,
d = (1/2) * 10.3 m/s^2 * (1.15 s)^2,
d = 5.955 m.

So, before attaining his full speed, the sprinter covered approximately 5.96 meters. That's just a couple of small steps for him!

(b) What was his final time for the race?

To find the final time, we need to calculate the time it takes to cover the remaining distance at a constant speed. We already know the distance covered during acceleration, so we subtract that from the total distance to find the remaining distance.

Remaining distance = 100 m - 5.96 m = 94.04 m.

Since speed = distance / time, we rearrange it to find the time:

time = distance / speed,
time = 94.04 m / 10.3 m/s,
time = 9.136 seconds.

Therefore, the sprinter's final time for the race is approximately 9.14 seconds. He didn't just sprint, he sprinted like a pro clown!

To solve this problem, we can use the equations of motion, specifically the equation for displacement and the equation for final velocity.

(a) To find the distance the sprinter ran before attaining his full speed, we need to determine the displacement during the acceleration phase.

The equation to calculate displacement is:

s = ut + (1/2)at^2

Where:
s = displacement
u = initial velocity
a = acceleration
t = time

In this case, the initial velocity (u) is 0 m/s as the sprinter starts from rest, the acceleration (a) can be found using the final velocity (10.3 m/s) and the time taken (1.15 seconds).

Rearranging the equation, we have:

s = (1/2)at^2

Substituting the values:

s = (1/2)(10.3 m/s)(1.15 s)^2

Calculating this expression gives us the answer to our first question.

(b) To find the final time for the race, we need to consider both the acceleration phase and the constant speed phase.

The time taken during the acceleration phase is already given as 1.15 seconds.

During the constant speed phase, the distance traveled can be calculated using the formula:

distance = speed x time

The distance covered during the constant speed phase is the remaining distance of the race, which is 100 meters minus the distance covered during the acceleration phase.

The time taken during the constant speed phase can be found by dividing the remaining distance by the speed (10.3 m/s):

time = distance / speed

Adding the time taken during the acceleration phase to the time taken during the constant speed phase will give us the final time for the race.

Let's calculate both the distance covered during the acceleration phase and the final time for the race.

(a) The distance covered during the acceleration phase:
s = (1/2)(10.3 m/s)(1.15 s)^2

(b) The remaining distance and the final time for the race:
distance = 100 m - [the distance covered during the acceleration phase]
time = [the remaining distance] / 10.3 m/s

By evaluating these expressions, you'll find the answers to both parts (a) and (b) of the question.

(a)

acceleration=(final velocity-initial velocity)/time

distance = (1/2)*(acceleration)*(time)^2 + (initial velocity)*(time)

(b) 100m - _____ (<-answer from (a))

find time using this formula:

distance = (1/2)*(acceleration)*(time)^2 + (initial velocity)*(time)

finally, add that time with 1.15 seconds