Total distance traveled with calculus

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Homework Help Overview

The problem involves analyzing the motion of a particle described by the equation s = t² - 5t + 6. Participants are tasked with finding the velocity and acceleration at specific times, as well as determining the total distance traveled after three seconds.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the calculation of velocity and acceleration, with one questioning the relationship between displacement and distance covered. There is exploration of the particle's position at various times and the implications of having zero displacement at two different instances despite non-zero velocity.

Discussion Status

The discussion is ongoing, with participants providing insights into the nature of the motion and the relationship between displacement and distance. Some guidance has been offered regarding the interpretation of the particle's motion and the significance of changing velocity.

Contextual Notes

Participants are navigating the complexities of motion as described by the given equation, questioning assumptions about displacement and distance. There is mention of a provided answer from a book, which adds to the uncertainty regarding the calculations.

albertrichardf
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Homework Statement


Suppose a particle responds to this equation of motion: s = t2 - 5t + 6
a) find the velocity at two seconds and three seconds
b) find the acceleration
c) find the total distance covered after 3 seconds

Homework Equations


s = t2 - 5t + 6
v = ds/dt = 2t - 5
a = dv/dt = 2

The Attempt at a Solution


ai) 2t - 5 = 4 - 5 = -1
aii) 2t - 5 = 6 - 5 = 1
(everything is fine here)

b) a = 2

c) The equation given above is for position, since it uses velocity. Is it actually possible to find the distance covered from the displacement?
I also tried finding the position at t = 1, t = 2 and t = 3 to find the distance.
at t = 1, s = 2
at t = 2, s = 4 - 10 + 6 = 0
at t = 3, s = 9 - 15 + 6 = 0
How is it possible that the displacement is 0 twice consecutively if the velocity is non-zero? Isn't that impossible?

I was actually going to calculate the displacement each time then add all together. The book had also already given an answer: 6.5m, although I'm not sure of how they got that.

Thanks for any answers.
 
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Look at the graph s=t^2 - 5t + 6
 
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Albertrichardf said:
How is it possible that the displacement is 0 twice consecutively if the velocity is non-zero? Isn't that impossible?
The velocity is non-zero but it's direction changed
 
Start from t=0 and keep in mind that the velocity is changing so the object will go one way, stop, and come back covering the same ground again explaining why the position at t=2 seconds coincides with the position at t = 3 seconds.
 
Albertrichardf; said:
Is it actually possible to find the distance covered from the displacement?
I also tried finding the position at t = 1, t = 2 and t = 3 to find the distance
Yes, distance through whch the wheels have turned can be found by adding up a number of individual displacements. But you don't calculate those separate displacements by dividing the trip according to some arbitrary time interval chosen because of convenience to you. The necessary time intervals are dictated by some characteristic of the motion itself.
 
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