Question #4146

A world class sprinter can reach top speed, 11.5m/s, in the first 15.0m of a race. What is the average deceleration of this sprinter and how long does it take her to reach that speed?

Expert's answer

Using the equations of the motion, we can write

h = v_{0}t+ at^{2}/2, where h is the covered distance, h_{0} is the initial position, v_{0} is the initial speed, a is the acceleration, t is the time.

Since h_{0} = 0, v_{0} = 0,

h = at^{2}/2.

t^{2} = 2h/a.

t = √2h/a

v = v0 + at = v0 + a*√2h/a = √2ha

v = √2ha

v^{2} = 2ha

a = v^{2}/2h

a = 11.5^{2}/(2*15) = 4.4 m/s^{2}.

t = √(2*15/4.4) = 2.61 s.

h = v

Since h

h = at

t

t = √2h/a

v = v0 + at = v0 + a*√2h/a = √2ha

v = √2ha

v

a = v

a = 11.5

t = √(2*15/4.4) = 2.61 s.

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