Highest velocity of a girl running a 60m dash

[HareJare]

Homework Statement

A girl is running 60m on 8.6 s. What is her biggest velocity, if the acceleration is konstant for the first 15m and then the velocity is constant after that?

Relevant Equations

v^2 - v0^2 = 2as
s = v * t

So i know the total time is 8.6 s

t1 + t2 = 8.6s

And from the (v^2 - v0^2 = 2as) i will get that after 15m the velocity which is also the highest velocity
v^2 = 30a

And then after that the velocity will be constant (s = v*t)
45/t2

Now i am stuck here because i have 3 equations but 4 variabels how do i solve this or is this even the right way to think about this problem?

 

[PeroK]

Homework Statement:: A girl is running 60m on 8.6 s. What is her biggest velocity, if the acceleration is konstant for the first 15m and then the velocity is constant after that?
Relevant Equations:: v^2 - v0^2 = 2as
s = v * t

So i know the total time is 8.6 s

t1 + t2 = 8.6s

And from the (v^2 - v0^2 = 2as) i will get that after 15m the velocity which is also the highest velocity
v^2 = 30a

And then after that the velocity will be constant (s = v*t)
45/t2

Now i am stuck here because i have 3 equations but 4 variabels how do i solve this or is this even the right way to think about this problem?

There is only one unknown, which is the constant acceleration $a$. Can you express the total time to cover $60m = 15m + 45m$ in terms of $a$?

 

[HareJare]

8.6 = sqrt(30/a) + ?

i don't know how to do the second term

 

[PeroK]

8.6 = sqrt(30/a) + ?

i don't know how to do the second term

You know the distance and you know the speed (in terms of $a$). Can you use that?

 

[PeroK]

PS It might be better to do everything in terms of the speed $v$ attained after $15m$.

You don't need to find the acceleration at all.

 

[HareJare]

PS It might be better to do everything in terms of the speed $v$ attained after $15m$.

You don't need to find the acceleration at all.

What do you mean?

 

[PeroK]

What do you mean?

Forget the acceleration and just focus on the maximum speed $v$.

As an alternative, note that the area under a velocity against time graph is the distance. If you draw a graph, there is an extremely simple geometric solution.

 

[HareJare]

Forget the acceleration and just focus on the maximum speed $v$.

As an alternative, note that the area under a velocity against time graph is the distance. If you draw a graph, there is an extremely simple geometric solution.

how do i get the time for the last 45m?

 

[PeroK]

how do i get the time for the last 45m?

You don't need to calculate that. Draw a velocity vs time graph.

 

[HareJare]

Okay i did that and now calculate the area?

 

[PeroK]

Okay i did that and now calculate the area?

You know the length of the base (that's $T = 8.6s$) and you know the height is $v$. The area of the big rectangle is $vT$. Can you relate that to the distances of $15m$ and $45m$ represneted by the triangle and the smaller rectangle?

 

[HareJare]

Okay thank you for your help but i don't understand how to do it this way.

 

[PeroK]

Okay thank you for your help but i don't understand how to do it this way.

You can, of course, do the same thing with the equations that represent the graph. Think about average speed for the first phase.