Solving Calc AP Problem 1: Total Distance 0-2

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SUMMARY

The total distance traveled by a particle moving along the x-axis from time t=0 to t=2 is calculated using the integral of the absolute value of its velocity function, v(t) = ln(t + 1) - 2t + 1. The correct formula for total distance is given by the integral from a to b of the absolute value of the velocity: ∫ab |v(t)| dt. This approach accounts for intervals where the particle moves backward, ensuring that the total distance is always a positive value.

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  • Understanding of calculus, specifically integration
  • Familiarity with velocity and displacement concepts
  • Knowledge of absolute value functions
  • Ability to graph functions and interpret their areas
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  • Learn how to compute definite integrals of absolute value functions
  • Study the properties of logarithmic functions, particularly ln(t + 1)
  • Explore particle motion problems involving velocity and acceleration
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Problem: A particle moves along the x-axis so that any time t >_ 0 its velocity is given by v(t) = ln(t + 1) - 2t + 1. What is the total distance traveled by the particle from t=0 to t=2?

Am I correct that the total distance is the area under the curve? I tried doing the integration on my calculator, and it gave me a negative answer. Then I graphed to make sure I didn't do anything wrong. I don't think I should be getting a negative answer, so... help please.
 
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I think the problem here is that your particle isn't always moving forward. When you measure total distance, you'll need to determine the intervals where the particle is moving backward (where velocity is negative) and take the absolute value of that distance.
Just a thought.
 
If it is velocity then area under the curve gives u displacement not distance for calculating distance apply the following formula

Distance covered from time t=a to t=b is

[tex]\int_a^{b} |v(t)|dt[/tex]

Or draw the graph of |v(t)| from the graph of v(t)

Area under |v(t)| will give u distance
 

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