A sprinter practicing for the 200-m dash accelerates uniformly from rest at A and reaches a top speed of 35 km/h at the 67-m mark. He then maintains this speed for the next 88 meters before uniformly slowing to a final speed of 32 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

0.705 m/s²

Explanation:

a) The sprinter accelerates uniformly from rest and reaches a top speed of 35 km/h at the 67-m mark.

Using newton's law of motion:

v² = u² + 2as

v = final velocity = 35 km/h = 9.72 m/s, u = initial velocity = 0 km/h,  s = distance = 67 m

9.72² = 0² + 2a(67)

134a = 94.484

a = 0.705 m/s²

b) The sprinter maintains this speed of 35 km/h for the next 88 meters. Therefore:

v = 35 km/h = 9.72 m/s, u = 35 km/h = 9.72 m/s, s = 88 m

v² = u² + 2as

9.72² = 9.72² + 2a(88)

176a = 9.72² - 9.72²

a = 0

c) During the last distance, the speed slows down from 35 km/h to 32 km/h.

u = 35 km/h = 9.72 m/s, v = 32 km/h = 8.89 m/s, s = 200 - (67 + 88) = 45 m

v² = u² + 2as

8.89² = 9.72² + 2a(45)

90a = 8.89² - 9.72²

90a = -15.4463

a = -0.1716 m/s²

The maximum acceleration is 0.705 m/s² which is from 0 to 67 m mark.

pecialized keyboards

the tires are rotating at 0.434 rpmlinear speed=45 m/sdiameter=0.550 mradius=0.275 mnow v= rωω= v/rω=45/0.275ω=163.6 rad/sω=163.6 rad/s (1rev/2π rad) (60s/1m)ω=1563 rpm