# Finding the instantaneous velocity

1. Aug 13, 2009

### frequin

1. The problem statement, all variables and given/known data

The questions refer to the attached graph

Q1 - calculate the instantaneous velocity of the object at time 2.5s

Q2 - calculate the instantaneous velocity of the object at time 7.5s

Q3- calculate the average velocity over the time of 7.5s

Q4- Is the motion of the object uniform or non-uniform?

2. Relevant equations

Unsure

3. The attempt at a solution

ANS 1: At 2.5s the object has traveled 4m, I am asumeing this means that at this time the instantaneous velocity is 4m per 2.5 sec witch then needs to be written in ms or kph?

ANS 2: same theory as above?

ANS 3: At 7.5 seconds the object has traveled 18.5m, to find the average I assume I divide the distance by time? but does this give me average speed not velocity? :S

ANS 4: I am saying non-uniform because the graph is not straight so the velocity and varies over the time

#### Attached Files:

• ###### physics.jpg
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2. Aug 13, 2009

### Pupil

Relevant equations
$$v = at + v_0$$

Does that help any? I can't see your graph due to the pending approval thing :grumpy:.

3. Aug 13, 2009

### frequin

sorry thats doesnt really help :S maybe you could explain what variables the letters represent?

here is a link to the graph

http://www.appendixj.org.au/mystuff/physics.jpg [Broken]

Last edited by a moderator: May 4, 2017
4. Aug 13, 2009

### Pupil

The a stands for acceleration, t for time and v_0 for initial velocity, but from the looks of your graph, it's irrelevant. Anyway...

Your answer to Q1 isn't quite right. The instantaneous velocity at any time is the slope of the tangent line at that time. It seems you only have a (rather poor) hand drawn graph, so you'll essentially be guessing at what the tangent is, but just taking the total distance up to that time and dividing by the time took is average velocity, not instantaneous velocity.

Last edited by a moderator: May 4, 2017
5. Aug 13, 2009

### RoyalCat

Manually take the derivative.

To take a derivative at a point, you take a secant through the point and one very close to it and then find the slope of the secant, which is approximately the same as the slope of tangent line at that point.

When you take the limit where the distance between the two points reaches 0, you get the actual derivative at that point, the slope of the tangent line.

Physically, you can approximate it by taking a ruler and just measuring $$\frac{\Delta y}{\Delta x}$$ and then take the limit as $$\Delta x \rightarrow 0$$

Last edited: Aug 13, 2009
6. Aug 13, 2009

### frequin

ok so here is my attempt at the first part of the question, please correct me if im wrong!

#### Attached Files:

• ###### velocity.jpg
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10.4 KB
Views:
99
7. Aug 14, 2009

### frequin

anyone?

8. Aug 14, 2009

### rock.freak667

it looks correct at first glance.