# Motion of centre of mass

1. Sep 30, 2014

### dk_ch

1. The problem statement, all variables and given/known data
The accompanying graph of position x versus time t represents the motion of a particle. If p and q are both
positive constants, the expression that best describes the acceleration  of the particle is

2. Relevant equations
(A) a =  - p - qt (B) a = p + qt (C) a = p + qt (D) a = p - qt
3. The attempt at a solution
In a given solution (D) is the correct answer.

But if we integrate the differential eq obtained from a = p-qt , we get velocity v = pt -0.5qt^2+c
the value of v ie dx/dt =0 for the inflexion pts on x vs t curve and we can get two values of t of the quadratic equation qt^2 -2pt -2c =0 but this gives one negative value of t which is impossible.
Then is (D) option correct? what is my wrong approach ?

2. Sep 30, 2014

### Staff: Mentor

The acceleration is zero at the inflection point, not the velocity.

Chet

3. Sep 30, 2014

### dk_ch

the plot was x-t . why not v=dx/dt = 0 at inflection pts?

4. Sep 30, 2014

### Staff: Mentor

An inflection point is one in which the second derivative is zero. On an x-t plot, the velocity is maximum (or minimum) at an inflection point. Take a look at the x vs t plot. Does it really look to you like dx/dt = 0 at the inflection point?

Chet

5. Sep 30, 2014

### Orodruin

Staff Emeritus
Because that is not the definition of an inflection point but that of a local extrema. The definition of an inflection point is that the curvature changes sign. This means that the second derivative should be zero and that the lowest non-vanishing derivative after that must be an odd derivative (ie, third derivative, fifth derivative, etc). But back to your actual question.

Normalizing with q = 1, we would get
$$t = p \pm \sqrt{p^2 + c}.$$
This does not mean that one of the roots is negative. If you look at your graph again, what is the sign of c? (First of all: What is the interpretation of c?)

6. Sep 30, 2014

### dk_ch

Actually I mean stationary inflection pts where first derivative is zero and in the graph there are two points where dx/dt is zero.. if t =0 then c = v ,at t=0, dx/dt is negative,hence c also. THANKS ALL for help.

Last edited: Sep 30, 2014
7. Sep 30, 2014

### Orodruin

Staff Emeritus
Well, you do not have any saddle points in the graph ... What you have is a minimum and a maximum, i.e., just two stationary points.

Back to the c issue. What does c describe? (Hint: What happens in your equations when t = 0?)

8. Sep 30, 2014

### nasu

You can answer in a more qualitative way. You can see from the graph that the second derivative is positive in the beginning (concave upwards) and negative in the end.
This means acceleration starting positive and ending negative. Only one answer satisfies this.

9. Oct 1, 2014

### dk_ch

thanks

10. Oct 1, 2014

thanks