Respuesta :
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:
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
