What Is the Difference Between Average and Instantaneous Velocity?

  • Thread starter Thread starter phragg
  • Start date Start date
  • Tags Tags
    Velocity
Click For Summary
SUMMARY

The discussion focuses on the concepts of average and instantaneous velocity as applied to a position-time graph. The average velocity was calculated as vave = -2 m/s over the interval from t = 1.50 s to t = 4.00 s, using the formula vave = Δx/Δt. The instantaneous velocity at t = 2.00 s was determined by finding the slope of the tangent line at that point, which was calculated as m = -3. The discussion also hints at identifying when the velocity is zero, which relates to the slope of the graph being horizontal.

PREREQUISITES
  • Understanding of position-time graphs
  • Knowledge of average velocity calculation
  • Familiarity with instantaneous velocity and tangent lines
  • Basic algebra for slope calculation
NEXT STEPS
  • Study the concept of derivatives in calculus to understand instantaneous velocity better
  • Learn about the graphical interpretation of velocity and acceleration
  • Explore the relationship between position, velocity, and time in physics
  • Investigate the conditions under which velocity is zero on a position-time graph
USEFUL FOR

Students studying physics, particularly those focusing on kinematics, as well as educators looking to explain the differences between average and instantaneous velocity.

phragg
Messages
2
Reaction score
0
A position-time graph for a particle moving along the x-axis is shown below:

(a) Find the average velocity in the time interval t = 1.50 s to t = 4.00 s.
(b) Determine the instantaneous velocity at t = 2.00 s by measuring the slope of the tangent line shown in the graph.
(c) At what value of t is the velocity zero?


http://img241.imageshack.us/img241/1514/tangetsb6.jpg

I figured that (a) was the easy part as I went ahead to solving that
v_{ave} = \frac{\Delta x}{\Delta t}

\Delta t = t_f - t_i
\Delta t = 4.00 s - 1.50s
\Delta t = 2.50 s

\Delta x = x_f - x_i
\Delta x = 2m - 7m
\Delta x = -5m

v_{ave} = \frac{-5m}{2.50s}
v_{ave} = -2m/s

So after that was done I went on to part (b) which first asked to find the slope of the tangent point was easily done by:

m = \frac{y_2 - y_1}{x_2 - x_1}
m = \frac{0 - 12}{4 - 0}
m = \frac{-3}{1}

Now I am completely stumped as to what they're asking for how to Determine the instantaneous Velocity at t = 2.00 s and Don't get me started on (c). Oh an p.s. Hey I'm new to PF ;p!
 
Last edited by a moderator:
Physics news on Phys.org
Hi, welcome to PF.
Actually, you already solved b :)
The key thing you are missing, is that instantaneous velocity at some time is the slope of the (t, x) graph at that point.

This should also help you solve c.
 

Similar threads

Replies
5
Views
2K
  • · Replies 9 ·
Replies
9
Views
1K
  • · Replies 6 ·
Replies
6
Views
1K
Replies
7
Views
2K
Replies
11
Views
3K
Replies
8
Views
1K
Replies
23
Views
2K
Replies
11
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
13
Views
2K