What is the Best Expression for the Acceleration of a Moving Particle?

[dk_ch]

Homework Statement

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

Homework Equations

(A) a = – - p -– qt (B) a = –p + qt (C) a = p + qt (D) a = p –- qt

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 ?[/B]

 

[Chestermiller]

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

Chet

 

[dk_ch]

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

Chet

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

 

[Chestermiller]

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

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

 

[Orodruin]

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.

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.

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?)

 

[dk_ch]

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?)

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.

• if f′(x) is zero, the point is a stationary point of inflection(http://en.wikipedia.org/wiki/Stationary_point), also known as a saddle-point

 

[Orodruin]

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?)

 

[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.

 

[dk_ch]

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.

thanks

 

[dk_ch]

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.

thanks