# Instananeous accerlerations on a velocity vs time graph

1. Jun 10, 2010

### mandy9008

1. The problem statement, all variables and given/known data
Find the instantaneous accelerations at 2.0 s, 10 s, and 18 s.

2. Relevant equations
a= lim (change in x) / (change in y)
...t->0

3. The attempt at a solution
2.0 s: 0 m/s2
18.0 s: 0 m/s2
10.0 s: ?
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

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2. Jun 10, 2010

### Staff: Mentor

Hint: Another way to express this is to say that acceleration is the slope of the velocity vs time graph.

3. Jun 12, 2010

### Axiom17

As Doc Al said, a good way to go about understanding this question is to simply think of the slope of the velocty vs. time graph as representing the acceleration. So the greater the gradient of the slope, i.e. the steaper the line, then the greater the acceleration at that time.

OK so you've got the graph plotted of what's going on, that's good. It's much easier to visualise what's going on and know if you're right when able to look at a graph.

This is stating, rather more confusingly than it could be, that as has been said already that the acceleration is equal to the change in x, i.e. the change in the velocity, with respect to the change in y, i.e. the change in time. Hopefully that's clearer now if it wasn't already.

OK so you've got the first 2 correct, not sure if that was through applying the formula, a bit of guesswork, or just intuition. They're simple once you understand what's going on here.

For the acceleration at $10.0s$, just apply the formula, using the value for the velocity at this time that can be read off the graph.

Hope that helps now

4. Jun 12, 2010

### Axiom17

Apologies just realised that method won't work very well I guess I'm not too with it at the moment!

Because the gradient of the line doesn't change around the time $10.0s$ what you can do is pick a point just before or after when $v \neq 0$ and thus you can get a non-zero value for the velocity and an accompanying time, then can find the acceleration using that simple formula.

5. Jun 13, 2010

### inky

I wonder I don't know who delete my calculation .

6. Jun 13, 2010

### Axiom17

.. pardon?!

7. Jun 13, 2010

### Redbelly98

Staff Emeritus
Your calculation was deleted because you are not supposed to do the calculation. The student who asked the question is supposed to do the calculation.

Did you not see the message I sent you?