2.1.2 AP calculus Exan particle move along the x-axis

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Discussion Overview

The discussion revolves around determining the total distance traveled by a particle moving along the x-axis, given its velocity function \( v(t) = 6t - t^2 \). Participants explore the relationship between velocity, position, and acceleration, considering the time interval from \( t = 0 \) to \( t = 3 \). The discussion includes elements of calculus, specifically integration and differentiation.

Discussion Character

  • Mathematical reasoning, Homework-related

Main Points Raised

  • One participant calculates the velocity at \( t = 3 \) to be 9, suggesting a straightforward evaluation of \( v(t) \) without deriving it from acceleration.
  • Another participant proposes a position function \( p(t) = 3t^2 - \frac{t^3}{3} + C \), indicating a potential approach to find the position from the velocity.
  • A participant questions whether the proposed function is for acceleration rather than position.
  • It is noted that the relationships between position, velocity, and acceleration are expressed as \( p'(t) = v(t) \) and \( v'(t) = a(t) \).
  • One participant states that the total distance can be calculated using the integral \( \int_{t_1}^{t_2} |v(t)| \, dt \), implying a method for determining total distance traveled.

Areas of Agreement / Disagreement

Participants have not reached a consensus on the total distance traveled, and multiple approaches and interpretations of the functions involved are present.

Contextual Notes

There are unresolved aspects regarding the integration limits and the necessity of considering the absolute value of velocity for total distance, as well as the constant \( C \) in the position function.

karush
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A particle moves along the x-axis. The velocity of the particle at time t is $6t - t^2$.
What is the total distance traveled by the particle from time $t = 0$ to $t = 3$
ok we are given $v(t)$ so we do not have to derive it from a(t) since the initial $t=0$ we just plug in the $t=3$ into $v(t)$ so
$6(3) - (3)^2=18-9=9$.
I thot I might of posted this earlier but I could not find it:rolleyes:
 
Last edited:
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Hmmm... $p(t)=3t^2-\frac{t^3}{3}+C$ (Wondering)
 
isn't that for a(t) ??
 
$p'(t)=v(t),\quad v'(t)=a(t)$
 
total distance = $\displaystyle \int_{t_1}^{t_2} |v(t)| \, dt$
 

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