Zero velocity for a time interval, what about acceleration?

[Monsterboy]

Homework Statement

For a particle moving in a straight line, if the velocity is zero for a time interval, the acceleration is zero at any instant within the time interval.

Relevant Equations

v = dx/dt
a = dv/dt

For a particle moving in a straight line, if the velocity is zero for a time interval, the acceleration is zero at any instant within the time interval.

I am told the above statement is true.

If I look at the equations

v = dx/dt
a = dv/dt

It looks like if the velocity is zero for a time interval, acceleration is also zero, but I am unable to visualize this in a long time interval.

Suppose a particle moves in the positive x direction from x1 to x2 and returns back to x1 taking a time interval of t2 - t1.

It's displacement is zero, so it's velocity will be zero for the time interval, right ?

But is the acceleration zero at any instant within the time interval ? For example, if we throw an object vertically upwards, it reaches some height and then falls back to the same initial height, so it's total displacement is zero because it comes back to the same point, but acceleration at any instant within the time interval is not zero, it is g. So, is the statement wrong ?

 

[BvU]

Suppose a particle moves

This is in contradiction with 'velocity is zero'
(because $dx$ is non-zero, and $v = {dx\over dt}$

$\$

 

[Monsterboy]

This is in contradiction with 'velocity is zero'
(because $dx$ is non-zero, and $v = {dx\over dt}$

$\$

Yea, but the question says velocity is zero "for a time interval", so what does it mean ?
Does it mean the particle moved somewhere and came back to the same position within the time interval or does it mean the particle never moved ?

 

[Orodruin]

Yea, but the question says velocity is zero "for a time interval", so what does it mean ?
Does it mean the particle moved somewhere and came back to the same position within the time interval or does it mean the particle never moved ?

Supposedly that $v(t) = 0$ for any $t$ inside the time interval.

 

[BvU]

velocity is zero "for a time interval", so what does it mean ?

It means that at all times within the interval $v \equiv {dx\over dt} = 0$

 

[Steve4Physics]

Homework Statement:: For a particle moving in a straight line, if the velocity is zero for a time interval, the acceleration is zero at any instant within the time interval.
Relevant Equations:: v = dx/dt
a = dv/dt

For a particle moving in a straight line, if the velocity is zero for a time interval, the acceleration is zero at any instant within the time interval.

I am told the above statement is true.

If I look at the equations

v = dx/dt
a = dv/dt

It looks like if the velocity is zero for a time interval, acceleration is also zero, but I am unable to visualize this in a long time interval.

Suppose a particle moves in the positive x direction from x1 to x2 and returns back to x1 taking a time interval of t2 - t1.

It's displacement is zero, so it's velocity will be zero for the time interval, right ?

But is the acceleration zero at any instant within the time interval ? For example, if we throw an object vertically upwards, it reaches some height and then falls back to the same initial height, so it's total displacement is zero because it comes back to the same point, but acceleration at any instant within the time interval is not zero, it is g. So, is the statement wrong ?

"For a particle moving in a straight line, if the velocity is zero for a time interval..."
means (in my opinion) that the velocity is zero during a finite time-interval (e.g. v=0 from t=5s to t=7s).

But it the particle is stationary during a finite time-interval, the reference to 'moving in a straight line' is not applicable (during the time-interval). So the original statement seems badly written.

 

[Lnewqban]

I leave my house in the morning and return to it every evening, but my car has been stopped for the times red lights have taken, accelerated, braked and moved at many different speeds.
Zero total displacement does not exclude velocity and acceleration.

The way I read the statement:
For a particle moving in a straight line between point A and point C, if the velocity is zero for a time interval the particle remains at point B, the acceleration is zero at any instant within the time interval the particle remains at point B.

Also:
For a particle moving in a straight line between point A and point D, if the velocity is constant for a time interval between the points B and C, the acceleration is zero at any instant within the time interval between the points B and C.

 

[robphy]

Suppose a particle moves in the positive x direction from x1 to x2 and returns back to x1 taking a time interval of t2 - t1.

It's displacement is zero, so it's velocity will be zero for the time interval, right ?

There needs to be a clarification.

Velocity should mean "instantaneous-velocity"
(which is defined at each instant of time)
whose magnitude is measured at each time by an idealized speedometer.

This should be distinguished from "average-velocity"
(which is defined for an interval of time, or a function of two times)
which is the constant velocity needed to travel between the endpoints in the same amount of time.
[Rant: it annoys me that "average-velocity" seems to be the typical textbook introduction of "velocity", possibly modeling the mathematics use of a sequence of the secant lines to get to the tangent line.
For physics, how does one have an average of something not yet defined?]

update:

Since there is likely
too much dependence on "technical sounding terms" (lacking or not using their proper technical definition)
and "formulas" (based on such "terms"),
I think
a position-vs-time graph would help visualize
instantaneous-velocity at an instant as the slope of the tangent line,
and
average-velocity between two times as the slope of the secant line, which connects the events $(t_1, x_1)$ and $(t_2, x_2)$.