A runner on a 100-m dash starts from rest and then accelerates to his maximum speed of 10.0 m/s to 4.00 s. He then continues with his speed until the end of the race.
a. How many seconds after he starts does the runner finish the race?
b. How far from the starting line is the runner at t = 4.00 s
c. What is the average speed of the runner for the entire race
d. If, upon the crossing the finish line, the runner immediately decelerates at a rate of 2.00 m/s^2, how far from the finish line will the runner stop?

- anonymous

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- anonymous

I am the physicist here

- anonymous

Do you want the full solutions or step by step walk you through?

- anonymous

Assuming his acceleration is constant:
\[a = \frac{ \left(10\frac{ m }{s}\right)-\left(0\frac{ m }{s}\right)}{4s - 0s} = 2.5 \frac{ m }{ s ^{2} }\]
Since his initial position and velocity was 0 m and 0 m/s respectively; by integrating the equation twice to obtain:
\[x(t) = \frac{ 1 }{ 2 } \left( 2.5 \frac{ m }{ s ^{2} }\right) t^{2}\]
Using this equation, we find the distance he traveled while accelerating by:
\[x(4s) = \frac{ 1 }{ 2 } \left( 2.5 \frac{ m }{ s ^{2} }\right) (4s)^{2} = 20m\]
He then proceeds to run at 10 m/s for the remaining 80m left in the sprint, which takes a further:
\[t = \frac{ 80m }{ 10 m s^{-1} } = 8 s\]
Therefore the total time for the dash is 12s.

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## More answers

- anonymous

You seem quite sound there my friend good job

- anonymous

Are you taking physics at uni? I am 2nd year physics

- anonymous

Yea I actually recently finished my BS and have started a graduate program. I just found this site and am getting some practice at tutoring the basics basics since I am trying to get a job with the Princeton review to help float the grad school costs.

- anonymous

Awesome! So physics BS from Priceton university?

- anonymous

Right now I am double majoring in medicine and physics and I am pursuing to be a medical physicist however sophistication on either subject requires sacrifice of the other to some extent.

- anonymous

No unfortunately not that prestigious, FSU.

- anonymous

I assume you were pure physics major then?

- anonymous

Physics and mathematics

- anonymous

That's a mutualism on both.

- anonymous

Medicine sacrifices physics and whereas physics promotes it... in terms of logical reasoning

- anonymous

Can't stand the universal hatred for physics amongst people

- anonymous

Totally... hang on for a minute I am typing up the remainder for this guy :)

- anonymous

Hey I think sig figs is set to 3 so 12.0 s would be more appropriate?

- anonymous

I think it's intentional that set it that way... Otherwise I would take 100m as having 1 sig fig which would round the answer to 10.... but deliberate use of .00 implies against it

- anonymous

my prof used to always pick me on sig figs until they were stuck as nightmares every now and then

- anonymous

The average speed of the runner throughout the race is calculated as follows:
Given the time taken for completion of the race;12.0 seconds
Total distance covered;100m
100m/12.0s=8.33m/s
The average speed of the runner throughout the race is 8.33m/s.
*Note that this is a scalar quantity for the direction is unspecified.

- anonymous

Given the speed at which the runner crosses the line;10m/s
The rate of negative acceleration(be careful with the use of deceleration);-2.00m/s^2
The time taken for the runner to stop is calculated as follows
10m/s-2.00m/s^2(t)=0
-2.00m/s^2(t)=-10m/s
t=5.00s
Therefore the time taken for the runner to come to the state of rest is 5.00s given the sig figs of 3

- anonymous

To calculate the distance from the point at which negative acceleration was applied, we consider the initial velocity as well as the negative acceleration by which the initial velocity was reduced to zero. Therefore,
10m/s(5s)-0.5(2.00m/s^2)(5s)^2=50m-25m=25.0m
Therefore the distance covered following negative acceleration upon crossing the line is 25.0m

- anonymous

The last question is two fold...

- anonymous

The average speed of the sprinter can be found by dividing total distance by total time:
\[v_{avg} = \frac{ 100 m }{ 12s } = 8.33 \frac{ m }{s}\]
The last part isn't to bad.
Take the initial velocity to be:
\[v_{0} = 10 \frac{ m }{ s }\]
And consider the finish line to be the new 0 point (aka located at 0 m).
Since the runners acceleration is again constant, but has magnitude (negative because his velocity is decreasing):
\[a = -2.0\frac{ m }{ s^{2} }\]
Integrating this equation once gives:
\[v(t) = -a t + v_{0} \rightarrow t = \frac{v(t) - v_{0}}{a}\]
Since his final veloctiy is zero after this interval -> v(T) = 0
\[T = \frac{- v_{0}}{a} = 5s\]
Integrating the velocity equation one more time:
\[x(t) = \frac{ -1 }{ 2 } a t^{2} + v_{0}t\]
And plugging in 5s yields:
\[x(5s) = \frac{ 1 }{ 2 } a (5s)^{2} + v_{0}(5s) = -25m + 50m = 25m \]

- anonymous

Whew editing is tough in this thing it is quite annoying and time consuming :(

- anonymous

Oh you beat me to the punch :p good job

- anonymous

Math majors are impressive with equations... wow I am so impressed.

- anonymous

I reason with concepts but math majors like every bit of equations.. wow

- anonymous

Well dont be too impressed answers are what matter and we both got the same :D

- anonymous

I think sig figs be 3

- anonymous

otherwise 1... which would render most of our answers totally inaccurate(100m) part

- anonymous

Youre probably right I have been quite sloppy with those.... math on numbers seems almost foreign to me now jk

- anonymous

but like I said consistency with 3 implies against it

- anonymous

Yea I spent all my time trying to get those to look good and be formatted correctly didnt even think of sig figs tbh

- anonymous

Yeah your answer is pretty professional. It's a mathematician's physics.

- anonymous

Mine is classical old fart physicist's physics

- anonymous

JK

- anonymous

:D

- anonymous

Oh god. Thankyou!! Guyssss sorry took so long to reply. I'm at school lolol thanks again I'll read everything first

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