Understanding Zero Velocity and Non-Zero Acceleration: Explained

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Zero velocity can occur at a single instant while acceleration remains non-zero, as demonstrated by a ball thrown upwards that momentarily stops at its peak height. Throughout its ascent and descent, the ball experiences constant downward acceleration due to gravity, even when its velocity is zero at the top. This concept can also be illustrated with other scenarios, such as a mass in a harmonic oscillator at maximum extension, where velocity is zero but acceleration is at its peak. The confusion often arises from misunderstanding the relationship between velocity and acceleration, where acceleration represents the rate of change of velocity. Ultimately, it is possible to have instantaneous zero velocity alongside non-zero acceleration.
pttest
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I heard it would be possible to have zero velocity & non zero acceleration (I know the opposite situation where there is velocity (constant), but zero acceleration). Could anyone please give me a clue on this?

Thanks in advance
 
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Throw a ball upwards.
 
Hi pttest! :smile:

When you throw a ball vertically upwards, and catch it again on the way down, it has the same acceleration (g downward) the whole time, but zero velocity when it reaches the top. :wink:
 
The prior post documents the only practical type case I can think of at the moment. A similar situation without velocity reversal could be a boat accelerating against a river current...as observed from shore.

An abstract situation might be at the moment when a distant observer sees a spaceship accelerating towards him and when the velocity reaches the observed cosmic expansion speed at the location of the spaceship, the observer would measure zero velocity...

How about an accelerating plane being overtaken by another higher fixed speed plane: at the moment the speeds are equal, velocity would be zero.

but so what??
 
I don't understand.. Velocity is the rate of change in position. Acceleration is the rate of change in velocity.

If the velocity isn't changing how can the acceleration be anything other than 0?
 
Nabeshin said:
Throw a ball upwards.

uhm, no. throwing a ball upwards does not produce zero velocity and nonzero acceleration. throwing a ball upwards constantly accelerates it at -9.8 meters per second per second.
 
davidsnider is absolutely right.
 
thomasxc said:
davidsnider is absolutely right.

No, he's not. And neither are you.

However, here's your chance to prove yourself: If the acceleration is 0 at the top of the ball's trajectory and 9.8 m/s2 on the way back down, at what point in time, t, does the acceleration become nonzero and what is the acceleration at that point?
 
DavidSnider said:
I don't understand.. Velocity is the rate of change in position. Acceleration is the rate of change in velocity.

If the velocity isn't changing how can the acceleration be anything other than 0?

Er... the velocity is zero only for an instant. It continues to change!

The ball being thrown up has a constant acceleration, which is downwards. It's velocity is changing, but changing at a constant rate! At some point, for a moment, it attains zero velocity.

Zz.
 
  • #10
ZapperZ said:
Er... the velocity is zero only for an instant. It continues to change!

The ball being thrown up has a constant acceleration, which is downwards. It's velocity is changing, but changing at a constant rate! At some point, for a moment, it attains zero velocity.

Zz.

OK, you can have an instantaneous velocity of 0, but at that point isn't your instantaneous acceleration also 0?
 
  • #11
DavidSnider said:
OK, you can have an instantaneous velocity of 0, but at that point isn't your instantaneous acceleration also 0?

Why? The velocity is changing!. That's the definition of it having an acceleration, which is -g!

Zz.
 
  • #12
I'm not quite sure why this is a problem, since this is standard high school physics.

a = \frac{dv}{dt} = -g (using the convention that upwards is positive),
v = -g\int{dt}
v = -gt + v_0

where v_0 is the initial velocity, and we let this to be positive since it was tossed upwards.

Now PLOT that as a function of t. You'll see that as gt grows in value, v will become smaller, until at some point, -gt + v_0 is zero! But look at how this was derived. It was derived for a constant acceleration of -g! Throughout the whole motion, the acceleration is a constant!

Zz.
 
  • #13
Naty1 said:
The prior post documents the only practical type case I can think of at the moment. A similar situation without velocity reversal could be a boat accelerating against a river current...as observed from shore.

Try the oscillating motion of a mass in a simple mass-spring harmonic oscillator. At the maximum extension, the mass temporarily has a zero velocity, but the acceleration is maximum.

Zz.
 
  • #14
I understand now. Thanks.
 
  • #15
pttest said:
I heard it would be possible to have zero velocity & non zero acceleration (I know the opposite situation where there is velocity (constant), but zero acceleration). Could anyone please give me a clue on this?

Thanks in advance

The problem is simply the way in which you phrased this statement.

Observe:

pttest said:
I heard it would be possible to have zero instantanous velocity & non zero acceleration (I know the opposite situation where there is velocity (constant), but zero acceleration).
 
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  • #16
excuse me. Zz corrected what i was trying to say.l
 
  • #17
DavidSnider said:
OK, you can have an instantaneous velocity of 0, but at that point isn't your instantaneous acceleration also 0?

The problem here is that you have a fundamental misunderstanding of what acceleration means in relation to velocity. Consider a curve of velocity vs. time. At some time, t, the instantaneous velocity is zero. The acceleration is the slope of the velocity curve at time t. You are confusing a value on a curve with its slope.
 

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