Question

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

Answer 1

Answer:

The maximum acceleration occurs from the zero mark to the 74m mark with a magnitude equal to 1.01m/s²

Explanation:

The sprinter accelerates in two instances

1. From the 0m to the 74m mark

2. From the 159m (85 +74) mark to the finish line.

Acceleration is a vector quantity having both magnitude and direction.

For the first case,

Initial velocity is zero at the sprinter starts from rest,

Final velocity is v = 44km/h = 44×1000/3600m/s = 12.2m/s

From the equations for constant acceleration motion,

v² = u² + 2as

Where u = initial velocity, a = acceleration, and s = distance covered ,

12.2² = 0² + 2a×74

148.84 = 148a

a = 148.84/148 = 1.01m/s²

Second instance of acceleration,

The sprinter slows down from an initial speed of 12.2 m/s to 10m/s (36km/hr) from the 159m mark to the finish line or over a distance of 41m (200 – 159).

So using the same equation above with u = 12.2m/s and v= 10m/s we have

10² = 12.2² + 2a×41

100 = 148.84 + 82a

82a = 100 – 148.84

82a = –48.84

a = –48.84/82 = 0.59m/s²

So the maximum acceleration occurs from the zero mark to the 74m mark with a magnitude equal to 1.01m/s²

Answer 2

Answer:

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