Howard J. answered • 10/14/19

Principal Mech Engr with 35+ years' on-the-job physics experience

**The 100m dash can be run by the best sprinters in 10.0 s.**

A 65 kg sprinter accelerates uniformly for the first 50m to reach top speed, which he maintains for the remaining 50 meters.

a) What is the average horizontal component of force exerted on his feet by the ground during acceleration?

b) What is the speed of the sprinter over the last 50m of the race?

I'm showing every little step since there may be some people who want to see the math.

Let t_{1} be the time to cover the first 50 m and t_{2} be the time to cover the remaining 50 m.

a) His speed as a function of time v(t) over 0≤D≤50 m is

v(t)=v(0)+at and since he starts from rest, v(0)=0 and after t_{1} seconds v(t_{1})=at_{1}

His progress D(t)=D(0)+v(0)t+0.5at^{2}

We can set D(0)=0 and since v(0)=0, then after t_{1}

D(t_{1})=0.5at_{1}^{2}=50

hence t_{1}=(100/a)^{1/2}_{ }=10/a^{1/2} [1]

After t_{1} his speed is constant and his progress over time t_{2 }is then

D_{2}=50=v_{2}t_{2}

v_{2}t_{2}=50 m

Since v_{2}=v(t_{1})

(at_{1})t_{2}=50 [2]

at_{1}t_{2}=50 [2]

If we assume he/she is a "best sprinter" then

t_{1}+t_{2}=10 [3]

Three equations and three unknowns [1], [2], and [3]

[1] into [2]

a(10/a^{1/2})t_{2}=50 [2]

10a^{1/2}t_{2}=50

a^{1/2}t_{2}=5 [4]

[1] into [3]

10/a^{1/2}+t_{2}=10 [5]

Now it's two equations and two unknowns [4] and [5]

a^{1/2}t_{2}=5 [4]

10/a^{1/2}+t_{2}=10 [5]

t_{2}=5a^{-1/2} [4]

[4] into [5]

10/a^{1/2}+5a^{-1/2}=10

15a^{-1/2}=10

a^{-1/2}=2/3

a^{1/2}=3/2

a=1.5^{2}

a=2.25 m/sec^{2}

F=ma=(65 kg)(2.25 m/sec^{2})= **0.15 kN**

Part a) was the hard part. Part b) is easy. I'll let you do it.

Awni B.

Thank you!10/14/19