Proving Acceleration Constant Between Point A & B

[no_alone]

Homework Statement

Car start to drive from point A to point B in a straight line ,
The distance from A to B is S, the time the car drove is T
Prove that there is a point in the drive where the acceleration of the car does not lower in absolute value from \frac{4S}{T^{2}}

the car start the drive from speed 0 and end the drive in speed 0
I know this is physics related question but I have in in calculus 2

Homework Equations

I almost sure you have to solve the question with taylor polynomial

The Attempt at a Solution

When I try built taylor polynomial when f(x) is the speed around x0 = when the speed reach S/T that is the average speed , the speed must reach the average speed

f(X) = f(x0) + f'(c)(x-x0)
f(x) = S/T +f'(c)(x-x0)

Now I do this for x=0 and for x=T and I can get that if the x0 in [0,T/4] or in [3T/4,T] f'(c) that is the acceleration is above \frac{4S}{T^{2}}

When I try to build the polynomial when f(x) is the location and again around x0 = the time the car speed reach S/T, It does not seem to add up, Because I don't really know what value to give to f(x0)...
Thank you for the help.

 

[TheForumLord]

Well... Vadim is your teacher?

Do it this way:
x(t)= x(0)+v(0)*t + a(c1)*t^2/2 when 0<c1<t...
So we'll get: x(t)=a(c1)/2 * t^2
If t=T/2 -> x(T/2) =a(c1)*T^2 /8...

Around T we'll get:
x(t) = S + a(c2)/2 * (t-T)^2 ... Put T/2 as t... Put both things you've got in equality and you're done :)

 

[no_alone]

Isn't the internet small
Thank you very much..
This was my very first try, But I thought , Ha c1 isn't c2 each one has a diffrent area and I got confused and stooped going this way.
...
Thanks.

 

[TheForumLord]

The whole world is a small place to live in :)

Anyway, which other courses you took this semester?