A sprinter practicing for the 200-m dash accelerates uniformly from rest at A and reaches a top speed of 44 km/h at the 74-m mark. He then maintains this speed for the next 85 meters before uniformly slowing to a final speed of 36 km/h at the finish line. Determine the maximum horizontal acceleration which the sprinter experiences during the run. Where does this maximum acceleration value occur

Maximum acceleration will be occur in the first stage of 0m - 74m.

Explanation:

From the question, it talks about 3 stages of the 200m practice.

We are asked to find the stage of maximum acceleration.

Now, there was no acceleration in the second stage because he sprinted with a constant velocity. Acceleration is zero at constant velocity.

Thus, it's only the first and third stage that he used acceleration.

First Stage;

This stage is from the 0m - 74m point.

While the final velocity is 44km/h and initial velocity is 0m/s

So, let's make use of the 3rd equation of motion;

v² = u² + 2as

Where u = initial velocity

a = acceleration

s = distance covered

v = final velocity

Making acceleration (a) the subject, we have;

a = (v² - u²)/2s

Now, we have to convert the final velocity (v) from km/h to m/s.

36km/h = 10m/s

Therefore, 44km/h = (44×10)/36 = 12.22m/s

Plugging in the relevant values to obtain ;

a = (12.22² - 0²)/2(74) = 149.3284/148 = 1.009 m/s²

Third Stage;

The initial velocity in this stage will be the final velocity from the second stage. Since he maintained 12.22m/s from the first stage to the end of the 2nd stage,

Initial velocity here (u) = 12.22 m/s

Final velocity is given as, v = 36 km/h = 10m/s

Distance covered (s) = 200 - (85+74) = 41m

We have same available parameters as in the stage 1,thus let's use,

a = (v² - u²)/2s

But in this stage, he is experiencing deceleration because he is gradually coming to a stop.

Thus, acceleration will be negative, so we now have,

-a = (v² - u²)/2s

Plugging in the relevant values to obtain ;

-a = (10² - 12.22²)/(2 x 41)

-a = -49.3284/82

Negative will cancel out and

a = 0.6016 m/s²

Now, let's compare the acceleration gotten in stages 1 and 2.

The acceleration in stage 1 is higher than that for the 3rd stage.

Thus, maximum acceleration will be occur in the first stage

at positions a, g kinetic energy is maximum.

at position d, potential energy is maximum.

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