Calculating velocity and position from time variating force

Click For Summary
SUMMARY

The discussion focuses on calculating velocity and position from a time-varying force using Newton's second law, represented as f=ma. The acceleration function is defined as a(t)=(4/5t^2i-3/5tj), leading to the velocity function v(t)=4/5i(t^3/3+c1)-3/5j(t^2/2+c2) after integration. The constants c1 and c2 can be eliminated by applying initial conditions, specifically assuming the initial position is the origin and the initial velocity is zero. The necessity of defining initial velocity is emphasized to avoid excessive undefined variables in the final equations.

PREREQUISITES
  • Understanding of Newton's second law (f=ma)
  • Knowledge of integral calculus for calculating velocity and position
  • Familiarity with vector notation in physics
  • Ability to apply initial conditions in differential equations
NEXT STEPS
  • Study the application of initial conditions in solving differential equations
  • Explore vector calculus, particularly in the context of physics
  • Learn about the implications of time-varying forces on motion
  • Investigate advanced integration techniques for complex functions
USEFUL FOR

Students in physics or engineering, particularly those studying dynamics and motion under variable forces, as well as educators looking for examples of applying calculus in physical scenarios.

jeremyskate
Messages
6
Reaction score
0

Homework Statement


Capture.PNG

2. Homework Equations [/B]
v(t)=∫a(t)dt
r(t)=∫v(t)dt

The Attempt at a Solution


f=ma
a(t)=f/m
a(t)=(4/5t^2i-3/5tj)
(integrate)
v(t)=4/5i(t^3/3+c1)-3/5j(t^2/2+c2)
how can i get rid of the c1 c2?
 
Physics news on Phys.org
You use initial conditions to get rid of them.
One condition is that initial position is the origin.
Another initial condition is the initial velocity, which is not stated, but needs to be. Since they haven't stated it, the simplest approach is to assume it is zero.
 
v(0)=0 would make the most sense but since it isn't mentioned I also thought of leaving the constants and integrating them as well to get the position function, but the end result would have way too much undefined variables...
 
If you assume that the initial velocity is ##v_1\mathbf{i}+v_2\mathbf{j}## then you will get rid of c1 and c2 and instead have ##v_1,v_2## in all your answers. That shouldn't be too onerous.
 
erroneous?
 

Similar threads

  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 17 ·
Replies
17
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 25 ·
Replies
25
Views
2K
Replies
14
Views
3K
  • · Replies 38 ·
2
Replies
38
Views
4K
  • · Replies 21 ·
Replies
21
Views
1K