# Velocity & acceleration

1. Aug 1, 2009

### pttest

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?

2. Aug 1, 2009

### Nabeshin

Throw a ball upwards.

3. Aug 1, 2009

### tiny-tim

Hi pttest!

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.

4. Aug 1, 2009

### Naty1

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??

5. Aug 1, 2009

### DavidSnider

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?

6. Aug 1, 2009

### thomasxc

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.

7. Aug 1, 2009

### thomasxc

davidsnider is absolutely right.

8. Aug 1, 2009

### negitron

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?

9. Aug 1, 2009

### ZapperZ

Staff Emeritus
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. Aug 1, 2009

### DavidSnider

OK, you can have an instantaneous velocity of 0, but at that point isn't your instantaneous acceleration also 0?

11. Aug 1, 2009

### ZapperZ

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

Zz.

12. Aug 1, 2009

### ZapperZ

Staff Emeritus
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. Aug 1, 2009

### ZapperZ

Staff Emeritus
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. Aug 1, 2009

### DavidSnider

I understand now. Thanks.

15. Aug 1, 2009

### Cyrus

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

Observe:

Last edited: Aug 1, 2009
16. Aug 1, 2009

### thomasxc

excuse me. Zz corrected what i was trying to say.l

17. Aug 1, 2009

### Cyrus

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.