Building up to understand integrals/area under curve.

  • Thread starter Thread starter LearninDaMath
  • Start date Start date
  • Tags Tags
    Building Curve
Click For Summary
SUMMARY

The discussion centers on the correct representation of a particle's motion under constant acceleration of 1 meter per second squared. The equation for acceleration is confirmed as a(t) = 1, while the velocity graph is derived from the acceleration graph through differentiation. Participants clarify that solving the second-order differential equation d²y/dt² = 1 involves integrating the right-hand side twice, resulting in a function with two constants that can be determined using initial conditions. The distinction between the graphs of acceleration and velocity is emphasized.

PREREQUISITES
  • Understanding of basic calculus concepts, including derivatives and integrals.
  • Familiarity with second-order differential equations.
  • Knowledge of initial conditions in motion equations.
  • Ability to interpret graphical representations of motion (velocity and acceleration).
NEXT STEPS
  • Study the process of integrating functions to solve differential equations.
  • Learn about the relationship between acceleration, velocity, and position in physics.
  • Explore the concept of initial conditions and their role in determining constants in equations.
  • Investigate graphical analysis of motion, focusing on velocity and acceleration graphs.
USEFUL FOR

Students of calculus, physics enthusiasts, and anyone looking to deepen their understanding of motion under constant acceleration.

LearninDaMath
Messages
295
Reaction score
0

Homework Statement

A particle starts at rest, then accelerates at a constant rate of 1 meter per second squared.

Homework Equations

Perhaps a(t) = t

The Attempt at a Solution

I have a series of calculus-related questions based on this statement, but first, I just want to know if the equation above, a(t)=t, and the graph below correctly represents the statement, "A particle starts at rest, then accelerates at a constant rate of 1 meter per second squared."
integralgraph.png
EDIT:

Or, is the correct equation for the statement: a(t) = 1

and the graph for the statement:

integralgraph1.png
 
Last edited:
Physics news on Phys.org
Yes your second graph/equation is correct. Your first graph is the correct graph for the velocity of the particle. I don't know how much calculus you know already, but acceleration is the derivative of velocity, so you can get your second graph by differentiating the first.
 
What you are asking is to solve the second-order differential equation [itex]\displaystyle \frac{d^2 y}{dt^2}=1[/itex]. Luckily for you, solving this differential equation requires near to no work compared to other second-order equations.

What you do is to take the integral of the right hand side twice. You will end up with a function involving two constants because you are integrating twice. You can then find those constants giving the starting velocity and the starting position some values.

And about your graph, the second one represents the acceleration. What does the first one represent?
 

Similar threads

  • · Replies 15 ·
Replies
15
Views
2K
  • · Replies 8 ·
Replies
8
Views
5K
Replies
8
Views
1K
  • · Replies 2 ·
Replies
2
Views
4K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 16 ·
Replies
16
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 13 ·
Replies
13
Views
6K