What is the displacement of the particle

Click For Summary
SUMMARY

The displacement of a particle with a velocity function defined as v = (5.5 m/s4)t3 - (5.4 m/s) can be calculated by integrating the velocity function over the interval from t = 3.5s to t = 9.0s. The correct approach involves finding the anti-derivative of the velocity function, resulting in the position function x = (5.5/4)t4 - 5.4t + x0. Evaluating this position function at the specified limits and subtracting the results provides the displacement, while ensuring that the constant x0 cancels out. The rapid increase in speed due to the t3 term explains the initially perceived large displacement value.

PREREQUISITES
  • Understanding of calculus, specifically integration and differentiation
  • Familiarity with the concepts of velocity and displacement
  • Knowledge of anti-derivatives and limits of integration
  • Basic understanding of polynomial functions
NEXT STEPS
  • Study the process of finding anti-derivatives in calculus
  • Learn about the Fundamental Theorem of Calculus
  • Explore polynomial functions and their properties
  • Practice problems involving integration of velocity functions to find displacement
USEFUL FOR

Students and professionals in physics and engineering, particularly those studying kinematics and calculus, will benefit from this discussion on calculating displacement from a velocity function.

ffrpg
Messages
12
Reaction score
0
Here's my problem, The velocity of a particle is given by
v = (5.5 m/s^4)t^3 - (5.4 m/s), what is the displacement of the particle during the interval t=3.5s to t=9.0s? I don't even know if I started it off right. I recalled seeing something similar to this in calc, so I tried integrating the formula given using 3.5 as 'a' and 9.0 as 'b'. I ended up getting 80440.5 as my answer, but it seems a bit large. I just need to know how to approach the problem. Thanks
 
Last edited by a moderator:
Physics news on Phys.org


"a" and "b"? I don't understand what those are! Oh, you mean the limits of integration. That's exactly what you should do.

What you need to know is that velocity is the derivative of position so position is the anti-derivative of velocity- that is, the integral.

Since v= dx/dt= (5.5)t^3 - 5.4, x= (5.5/4)t^4- 5.4t+ x0. (x0 is the "unknown constant" and would be the position of the particle at t=0). Evaluate at t= 9.0s and t= 3.9s and subtract. (The x0 in each term will, of course, cancel).

Why does your answer seem "a bit large"? With the t^3 term the speed is going to increase rapidly!
 
Hey thanks for the help. I figured out what I did wrong. I integrated incorrrectly. THANKS!
 

Similar threads

Calculus Problem: acceleration, speed, and displacement of a particle
  • · Replies 1 ·
Replies
1
Views
2K
Relating acceleration to distance and time
  • · Replies 5 ·
Replies
5
Views
3K
Analyzing Motion: Deriving Displacement Graphs from First Principles
  • · Replies 6 ·
Replies
6
Views
2K
Needs clarity on this displacement vector diagram
  • · Replies 3 ·
Replies
3
Views
1K
Why do we use area under the curve to find displacement in particle motion?
  • · Replies 6 ·
Replies
6
Views
2K
Wave movement and particle movement
  • · Replies 23 ·
Replies
23
Views
1K
Net displacement of a particle given its equation of motion
  • · Replies 7 ·
Replies
7
Views
3K
Solving Vector Diagrams: Finding Displacements & Velocities
  • · Replies 5 ·
Replies
5
Views
2K
What is the smallest value of angular displacement of the raft?
  • · Replies 54 ·
2
Replies
54
Views
4K
How to know 'the displacement of a particle is 'negative'?
  • · Replies 6 ·
Replies
6
Views
1K