How to get velocity and position from an acceleration-time graph?

Click For Summary
SUMMARY

The discussion centers on calculating velocity and position from an acceleration-time graph for a particle with an initial position of 10 m and an initial velocity of 9 m/s. The correct formulas to use are v = v_o + at and x = v_ot + (1/2)at^2. The user initially calculated the velocity at 7 seconds as -12 m/s and the position as -0.5 m, which were incorrect. Proper application of the equations is necessary to derive the correct values for both velocity and position.

PREREQUISITES
  • Understanding of kinematic equations, specifically v = v_o + at and x = v_ot + (1/2)at^2.
  • Basic knowledge of acceleration, velocity, and position in physics.
  • Familiarity with graph interpretation, particularly acceleration-time graphs.
  • Ability to perform algebraic calculations involving time and constants.
NEXT STEPS
  • Review the application of kinematic equations in physics problems.
  • Practice solving problems involving acceleration-time graphs.
  • Learn how to derive velocity and position from different types of motion graphs.
  • Explore online physics simulation tools to visualize motion and graph relationships.
USEFUL FOR

Students studying physics, educators teaching kinematics, and anyone seeking to improve their understanding of motion analysis using graphs.

veryconfused
Messages
2
Reaction score
0
Consider the plot below describing the acceleration of a particle along a straight line with an initial position of 10 m and an initial velocity of 9 m/s.

http://img93.imageshack.us/img93/5492/grapho.jpg

1. What is the velocity at 7 s?
2. What is the position at 7 s?
 
Last edited by a moderator:
Physics news on Phys.org
Please help me! I got the answer -12 for #1 and -0.5 for #2, but the computer's telling me these are wrong (it's an online worksheet). I don't understand why?
 
Well you can use the two default equations.
[tex]v=v_o +at[/tex]
[tex]x=v_ot+\frac{1}{2}at^2[/tex]
You'll have to use each equation twice for each problem though.
 

Similar threads

  • · Replies 13 ·
Replies
13
Views
2K
  • · Replies 2 ·
Replies
2
Views
5K
Replies
1
Views
3K
  • · Replies 4 ·
Replies
4
Views
4K
  • · Replies 13 ·
Replies
13
Views
4K
  • · Replies 3 ·
Replies
3
Views
6K
Replies
12
Views
2K
  • · Replies 6 ·
Replies
6
Views
3K
Replies
6
Views
2K
  • · Replies 13 ·
Replies
13
Views
1K