Proving Acceleration Constant Between Point A & B

In summary, the conversation is about solving a physics-related question using calculus. The goal is to prove that there is a point in the drive where the acceleration of the car does not lower in absolute value from 4S/T^2. The solution involves using Taylor polynomials and setting up equations for the speed and location of the car at different points in time. The conversation also briefly touches on other courses the participants have taken during the semester.
  • #1
no_alone
32
0

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 [tex]\frac{4S}{T^{2}}[/tex]

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 [tex]\frac{4S}{T^{2}}[/tex]

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.
 
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  • #2


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 :)
 
  • #3


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.
 
  • #4


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

Anyway, which other courses you took this semester?
 

Related to Proving Acceleration Constant Between Point A & B

1. What is acceleration and how is it measured?

Acceleration is the rate of change of an object's velocity over time. It is measured in meters per second squared (m/s^2) or feet per second squared (ft/s^2). It can be calculated by dividing the change in velocity by the change in time.

2. How do you prove that acceleration is constant between two points?

To prove that acceleration is constant between two points A and B, you must measure the object's velocity at both points and calculate the change in velocity over the change in time. If the resulting value is the same, then the acceleration is constant between the two points.

3. What are the units of acceleration?

The units of acceleration are meters per second squared (m/s^2) or feet per second squared (ft/s^2).

4. What factors can affect the constant acceleration between two points?

The constant acceleration between two points can be affected by external forces such as friction, air resistance, and gravity. These forces can cause the object's velocity to change, resulting in a non-constant acceleration.

5. How is acceleration related to Newton's Second Law of Motion?

Newton's Second Law of Motion states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. In other words, the greater the force applied, the greater the acceleration, and the greater the mass, the smaller the acceleration.

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