Simple math problem in one dimension

  • Thread starter Thread starter Slusho
  • Start date Start date
  • Tags Tags
    Dimension
Click For Summary
SUMMARY

The discussion focuses on a physics problem involving a particle subjected to a time-varying force F(t) = F0(1 - t/T). The particle's speed and distance traveled at time T are derived as v(T) = v0 + a0T/2 and x(T) = v0T + a0T²/3, where a0 = F0/m is the initial acceleration. The user attempts to apply Newton's second law and the relationships between acceleration, velocity, and position but struggles to derive the correct coefficients. The key takeaway is the importance of understanding the time derivatives of motion in solving such problems.

PREREQUISITES
  • Understanding of Newton's second law of motion
  • Familiarity with kinematic equations for linear motion
  • Basic calculus concepts, particularly derivatives
  • Knowledge of force, mass, and acceleration relationships
NEXT STEPS
  • Study the derivation of kinematic equations in one dimension
  • Learn about time-dependent forces and their effects on motion
  • Explore the relationship between acceleration, velocity, and position through calculus
  • Investigate examples of non-constant forces in physics problems
USEFUL FOR

Students of physics, particularly those studying mechanics, educators teaching kinematics, and anyone interested in applying calculus to motion problems.

Slusho
Messages
17
Reaction score
0

Homework Statement


A particle of mass m is subjected to a net force F(t) given by F(t)=F0(1-t/T)i; that is F(t) equals F0 at t=0 and decreases linearly to zero in time T. The particle passes the origin x=0 with velocity v0i. Show that at the instant t=T and F(t) vanishes, the speed v and distance x traveled are given by v(T)=v0+a0T/2, and x(T)=v0T+a0T2/3, where a0=F0/m is the initial acceleration. Compare these results with Eqs (vx=v0x+axt and x=x0+v0xt+1/2axt2).


Homework Equations


I should only need those listed in the problem


The Attempt at a Solution


F=ma=F0(1-t/T)
F0=ma0
a=a0(1-t/T)
a=a0-a0t/T

v=v0+at
v=v0+(a0-a0t/T)t

When t=T, those a0's cancel each other out and you end up with zero. I've also tried starting with the other one, and taking derivatives, to no avail. I can't get the math to work out to give me those coefficients.

Help is GREATLY appreciated.
 
Physics news on Phys.org
So you've already written Newton's 2nd Law, that's important. But what are the time derivative relationships between acceleration, velocity, and position? Use those relationships to work backwards from Newton's 2nd Law to obtain v(t) and x(t).
 

Similar threads

  • · Replies 9 ·
Replies
9
Views
2K
Replies
2
Views
4K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 10 ·
Replies
10
Views
2K
Replies
3
Views
2K
  • · Replies 29 ·
Replies
29
Views
3K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 4 ·
Replies
4
Views
1K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K