Net displacement of a particle given its equation of motion

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Homework Help Overview

The discussion revolves around determining the net displacement of a particle given its equation of motion, specifically the velocity function v=18-2t^2 m/s and an initial position of s_0 = -3 m. Participants are examining the calculations over a time interval of 5 seconds.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the use of the integral to find the position function s(t) and question the limits of integration. There are attempts to clarify the correct application of the formula and the impact of initial conditions on the results.

Discussion Status

Some participants have provided guidance on the correct formula and highlighted the importance of careful consideration of signs and units. There is ongoing clarification regarding the limits of integration and the interpretation of results, with no explicit consensus reached yet.

Contextual Notes

There are indications of confusion regarding the integration process and the correct upper limit for the integral, as well as the handling of initial conditions. Participants are also reflecting on the implications of their calculations for total distance and net displacement.

javii
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Homework Statement


Hello PF,

I need some help with the assignment given:
v=18-2t^2 m/s, where t is in second. When t= 0 the position of the particle is s_0 = - 3 m.
For the first 5 seconds.
Determine the total distance ( i got it to 65,33 ft) and the net displacement Δs, and the value of s at the end of the interval.

The Attempt at a Solution


I guess I have to use the formula:
s(t)=s_0 + ∫_0 ^t v(t) dt

s(t)=3+∫_0^3 18-2t^2 dt
 
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javii said:
s(t)=s_0 + ∫_0 ^t v(t) dt
That is the correct equation.
S0= -3m.
 
... and tf = 5 seconds.
 
Be careful with signs and units and integral borders. Apart from that, the total distance is fine, and the formula for s(t) will help with the net displacement Δs and s at the end.
 
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cnh1995 said:
That is the correct equation.
S0= -3m.
But when I integrate it, I do not get the correct answer.
 

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javii said:
But when I integrate it, I do not get the correct answer.
Why are you multiplying the integral by -3?
Also, the upper limit of the integral is not 3.
gneill said:
... and tf = 5 seconds.
 
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cnh1995 said:
Why are you multiplying the integral by -3?
Also, the upper limit of the integral is not 3.
May bad. So this should be correct?
 

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That number is relevant. Which part did you answer with it?
 

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