Displacement in the 5th second.

  • Thread starter Thread starter zorro
  • Start date Start date
  • Tags Tags
    Displacement
Click For Summary
SUMMARY

The discussion centers on calculating the distance covered by a particle in the fifth second of its motion, given an initial velocity of 9 m/s due east and a constant acceleration of 2 m/s² due west. The correct distance covered is determined to be 0.5 m, contrary to an initial incorrect assertion of 0 m. The participants clarify the distinction between displacement and distance traveled, ultimately arriving at the correct answer through a method involving velocity calculations and integration.

PREREQUISITES
  • Understanding of kinematic equations, specifically displacement formulas.
  • Knowledge of basic calculus, particularly integration of velocity functions.
  • Familiarity with concepts of velocity, acceleration, and their directional components.
  • Ability to differentiate between displacement and distance traveled in physics.
NEXT STEPS
  • Study kinematic equations for uniformly accelerated motion.
  • Learn about the principles of integration in physics, particularly for motion analysis.
  • Explore the differences between displacement and distance in various motion scenarios.
  • Practice problems involving velocity and acceleration to solidify understanding of these concepts.
USEFUL FOR

Students studying physics, particularly those focusing on kinematics, as well as educators looking for clarification on common misconceptions regarding displacement and distance in motion problems.

zorro
Messages
1,378
Reaction score
0

Homework Statement


A particle has an initial velocity of 9m/s due east and a constant acceleration of 2 m/s2 due west. The distance covered by the particle in the fifth second of its motion is?


The Attempt at a Solution



Using the formula for the displacement in the nth second,
Sn = u + a(n-0.5)

I got S5=0.

The answer given is 0.5m.
 
Physics news on Phys.org
The answer given is wrong.

ehild
 
thanks!
 
Actually, the answer in the book is correct.

You didn't answer the question asked -- you gave the answer to a similar but different question.

(hint: zero is obviously a wrong answer to the question asked)
 
Thanks Hurkyl! As it was "displacement" in the title, I did not recognise that "distance covered" was distance travelled, and mixed it with displacement. In this case, the book is right.

ehild
 
Last edited:
Damn! even I thought it is displacement. I got the answer now.
This is how I did it -
observe that the velocity at the beginning of the 5th second is 1m/s and at its end is -1m/s. So we have to find the displacement for half second by symmetry and double it to get the distance travelled.

s = 1/2 -1/2*2*1/4 = 1/4m
Hence total distance is 0.5m. Is there any other method?
 
The distance traveled is equal to the integral of the speed (magnitude of the velocity, |v|) with respect to time. v is positive till t=4.5 s, and negative afterwards, so the integral is split into ∫(9-2t)dt from 4 to 4.5 and ∫(2t-9)dt from 4.5 to 5.

ehild
 

Similar threads

  • · Replies 8 ·
Replies
8
Views
2K
Replies
12
Views
3K
Replies
17
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K
Replies
11
Views
7K
Replies
4
Views
2K
  • · Replies 27 ·
Replies
27
Views
3K
  • · Replies 6 ·
Replies
6
Views
3K
Replies
23
Views
2K