Total Distance travelled by a particle

  • Thread starter Thread starter phat2107
  • Start date Start date
  • Tags Tags
    Particle
Click For Summary
SUMMARY

The total distance traveled by a particle from time 0 to 10 is determined by the position function s(t) = 2t^3 - 15t^2 + 24t. The derivative s'(t) = 6t^2 - 30t + 24 indicates the particle's direction of movement. The particle moves right from t=0 to t=1, left from t=1 to t=4, and right again from t=4 to t=10. The total distance traveled is the sum of the absolute distances in each segment, resulting in a total distance of 60 units, despite a net displacement of 40 units.

PREREQUISITES
  • Understanding of calculus concepts, specifically derivatives
  • Familiarity with polynomial functions and their properties
  • Knowledge of motion along a straight line
  • Ability to calculate definite integrals for distance
NEXT STEPS
  • Study the Fundamental Theorem of Calculus for distance calculations
  • Learn about particle motion and its representation through position functions
  • Explore the concept of total distance versus net displacement
  • Practice solving polynomial equations and their derivatives
USEFUL FOR

Students studying calculus, physics enthusiasts, and anyone interested in understanding motion along a straight line and distance calculations.

phat2107
Messages
11
Reaction score
0

Homework Statement


what is the total distance traveled from time 0 - 10

the equations is 2t^3-15t^2+24t


Homework Equations





The Attempt at a Solution


time 0 it is at 0
time 10 its at 740

the answer is not 740

i have no idea how to solve this equation, no clue as to what I am supposed to do

any help will be great

thanks
 
Physics news on Phys.org
Does the equation given describe the path of the particle?... or does it describe something else?
 
"A particle moves along a straight line and its position at time t is given by s(t)= 2t^3 - 15 t^2 + 24 t"
 
phat2107 said:
"A particle moves along a straight line and its position at time t is given by s(t)= 2t^3 - 15 t^2 + 24 t"

So s'(t)= 6t^2- 30t+ 24= 6(t^2- 5t+ 4)= 6(t- 4)(t- 1). For t between 0 and 1, s'(t) is positive (both t-1 and t- 4 are negative) so the particle is moving to the right. Between t= 1 and t= 4, s'(t) is negative (t- 4 is still negative but t- 1 is now positive) so the particle is moving to the left. Between t= 4 and t= 10, s'(t) is positive (t- 4 and t- 1 are now both positive) so the particle is moving to the right. The "total distance" traveled is the distance traveled between t= 0 and 1 plus the positive distance traveled between t= 1 and t= 4 plust the positive distance traveled between t= 4 and t= 10.p

it's like going 20 miles to the east, then 10 miles back to west, the 30 miles to the east again.
You are only 20- 10+ 30= 40 miles from your starting point but your speedometer will say that you have gone 20+ 10+ 30= 60 miles.
 
Last edited by a moderator:
understood, the car example made it clear

thanks
 

Similar threads

  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 3 ·
Replies
3
Views
7K
  • · Replies 5 ·
Replies
5
Views
4K
Replies
7
Views
2K
  • · Replies 7 ·
Replies
7
Views
8K
  • · Replies 1 ·
Replies
1
Views
9K
  • · Replies 4 ·
Replies
4
Views
1K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 4 ·
Replies
4
Views
5K
  • · Replies 1 ·
Replies
1
Views
1K