# 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

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Answer:The maximum acceleration occurs from the zero mark to the 74m mark with a magnitude equal to 1.01m/s²Explanation:The sprinter accelerates in two instances1. From the 0m to the 74m mark2. 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 + 2asWhere u = initial velocity, a = acceleration, and s = distance covered ,12.2² = 0² + 2a×74148.84 = 148a a = 148.84/148 = 1.01m/s²Second 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×41100 = 148.84 + 82a82a = 100 – 148.84 82a = –48.84a = –48.84/82 = 0.59m/s²So 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/sSo, let's make use of the 3rd equation of motion;v² = u² + 2asWhere u = initial velocitya = accelerations = distance coveredv = final velocityMaking 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/sTherefore, 44km/h = (44×10)/36 = 12.22m/sPlugging in the relevant values to obtain ;a = (12.22² - 0²)/2(74) = 149.3284/148 = 1.009 m/s²Third 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/sFinal velocity is given as, v = 36 km/h = 10m/sDistance covered (s) = 200 - (85+74) = 41mWe 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/s²Now, 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