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

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