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 t1 be the time to cover the first 50 m and t2 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 t1 seconds v(t1)=at1
His progress D(t)=D(0)+v(0)t+0.5at2
We can set D(0)=0 and since v(0)=0, then after t1
D(t1)=0.5at12=50
hence t1=(100/a)1/2 =10/a1/2 [1]
After t1 his speed is constant and his progress over time t2 is then
D2=50=v2t2
v2t2=50 m
Since v2=v(t1)
(at1)t2=50 [2]
at1t2=50 [2]
If we assume he/she is a "best sprinter" then
t1+t2=10 [3]
Three equations and three unknowns [1], [2], and [3]
[1] into [2]
a(10/a1/2)t2=50 [2]
10a1/2t2=50
a1/2t2=5 [4]
[1] into [3]
10/a1/2+t2=10 [5]
Now it's two equations and two unknowns [4] and [5]
a1/2t2=5 [4]
10/a1/2+t2=10 [5]
t2=5a-1/2 [4]
[4] into [5]
10/a1/2+5a-1/2=10
15a-1/2=10
a-1/2=2/3
a1/2=3/2
a=1.52
a=2.25 m/sec2
F=ma=(65 kg)(2.25 m/sec2)= 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