A Graph analysis problem to find when the acceleration is zero

[uranium92]

Homework Statement

I have attached the question below in the image. It is not a homework but I am solving Introductory Mechanics by Morin.

Relevant Equations

$$v = \frac{dx}{dt}$$
$$a = \frac{dv}{dt}$$

I had deduced that B,E,H are the places where acceleration will be zero, but when I read the solutions it showed that K also has a = 0. It said it had maximum slope and then said a = 0. But I couldn't understand why? Please help.

 

[TSny]

It said it had maximum slope

What does the slope represent on the x vs t graph?

 

[neilparker62]

Looks like K is a point of inflection on the x(t) graph.

 

[hmmm27]

Think; what sort of behaviour is expected to represent $a=0$ in each graph.

acceleration vs time - not worth mentioning except to note that it's not worth mentioning.

velocity vs time - if there is no (ie:zero) acceleration/deceleration, what does that do to the velocity ? How will that be represented in the graph ? note that I disagree with your assertion that $E$ and $H$ represent zero acceleration

distance vs time - how would the velocity as determined by $a=0$ show up on a $d:t$ graph ? and I agree with the official answer concerning the locus around point $K$ as $a=0$

 

[uranium92]

What does the slope represent on the x vs t graph?

It represents the velocity in a x vs t graph.

 

[uranium92]

Looks like K is a point of inflection on the x(t) graph.

I don't know what inflection is. If you could tell me or give some links it would be nice.

 

[uranium92]

Think; what sort of behaviour is expected to represent $a=0$ in each graph.

acceleration vs time - not worth mentioning except to note that it's not worth mentioning.

velocity vs time - if there is no (ie:zero) acceleration/deceleration, what does that do to the velocity ? How will that be represented in the graph ? note that I disagree with your assertion that $E$ and $H$ represent zero acceleration

distance vs time - how would the velocity as determined by $a=0$ show up on a $d:t$ graph ? and I agree with the official answer concerning the locus around point $K$ as $a=0$

I deduced about E and H because the slopes at both points are zero or say parallel to the axis but what I don't understand is how at K acceleration = 0.

 

[TSny]

It represents the velocity in a x vs t graph.

Yes. And, as you said, the slope is maximum at point K.

Which point of the v vs t graph best corresponds to point K of the x vs t graph?

 

[hmmm27]

I deduced about E and H because the slopes at both points are zero or say parallel to the axis but what I don't understand is how at K acceleration = 0.

I see your point on the $v:t$ graph, though I still disagree with it : I'd argue that a constant velocity (ie: zero acceleration) is determined by a line not a point, and the lines don't look straight (and level) to me. YMMV.

Regarding $K$, don't look directly at the point ; look at the segment - what does a straight line in a $d:t$ graph represent in terms of velocity ?

 

[uranium92]

Yes. And, as you said, the slope is maximum at point K.

Which point of the v vs t graph best corresponds to point K of the x vs t graph?

Isn't it mentioned in the question that all the graphs are not related to each other?

 

[TSny]

Isn't it mentioned in the question that all the graphs are not related to each other?

Yes. But I still think my question is relevant.

 

[uranium92]

I see your point on the $v:t$ graph, though I still disagree with it : I'd argue that a constant velocity (ie: zero acceleration) is determined by a line not a point, and the lines don't look straight (and level) to me. YMMV.

Regarding $K$, don't look directly at the point ; look at the segment - what does a straight line in a $d:t$ graph represent in terms of velocity ?

Got it. Does it mean the velocity is constant which means zero acceleration?

 

[hmmm27]

Got it. Does it mean the velocity is constant which means zero acceleration?

A constant velocity is the definition of zero acceleration.

 

[uranium92]

A constant velocity is the definition of zero acceleration

Thank you very much for your help.

 

[TSny]

A constant velocity is the definition of zero acceleration.

No. You can have zero acceleration at an instant of time even though the velocity is not constant.

For example, consider the case where the velocity as a function of time is given by the simple equation $v = t^2$. Note that the velocity is always changing with time. There is no interval of time over which the velocity is constant.

For this velocity function the acceleration as a function of time is given by $a = 2t$.
At $t = 0$ the acceleration is zero even though the velocity is not constant.

So, zero acceleration at some instant of time does not imply constant velocity.

Likewise, at point K of the x vs t graph, the acceleration is zero. We can say that the velocity is maximum at K, but we can't conclude that the velocity is constant near K.

 

[uranium92]

No. You can have zero acceleration at an instant of time even though the velocity is not constant.

For example, consider the case where the velocity as a function of time is given by the simple equation $v = t^2$. Note that the velocity is always changing with time. There is no interval of time over which the velocity is constant.

For this velocity function the acceleration as a function of time is given by $a = 2t$.
At $t = 0$ the acceleration is zero even though the velocity is not constant.

So, zero acceleration at some instant of time does not imply constant velocity.

Likewise, at point K of the x vs t graph, the acceleration is zero. We can say that the velocity is maximum at K, but we can't conclude that the velocity is constant near K.

Understood it, Thanks for your help.

 

[neilparker62]

I don't know what inflection is. If you could tell me or give some links it would be nice.

A point of inflection of function f(x) occurs when the second derivative is zero - ie f''(x)=0. On the graph of x(t) in your problem, this corresponds to the point where the slope of the graph stops increasing and starts decreasing.

 

[uranium92]

A point of inflection of function f(x) occurs when the second derivative is zero - ie f''(x)=0. On the graph of x(t) in your problem, this corresponds to the point where the slope of the graph stops increasing and starts decreasing.

Thanks for explaining.

 

[Delta2]

No. You can have zero acceleration at an instant of time even though the velocity is not constant.

You are very right here, however i think @hmmm27 was saying that if acceleration is zero at all times then the velocity is constant.