Find the distance given a function for v(t)

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Discussion Overview

The discussion revolves around finding the distance traveled given a velocity function, v(t) = 2t + 5t^2. Participants explore the relationship between velocity and distance, particularly focusing on integrating the velocity function and addressing the implications of given speed values.

Discussion Character

  • Mathematical reasoning, Homework-related, Technical explanation

Main Points Raised

  • One participant presents the velocity function and derives the distance function through integration, noting the anti-derivative.
  • Another participant points out that the derived distance function is in terms of time, questioning the next steps to relate it to the given speed values.
  • A third participant expresses confusion about using velocity values directly to find distance, indicating a misunderstanding of the relationship between speed and time.
  • A later reply highlights a potential oversight in the initial equation, specifically regarding the coefficient in the velocity term, suggesting that it should be considered to find time values from the velocity equation.

Areas of Agreement / Disagreement

Participants do not reach a consensus on how to proceed with the problem, as there are differing interpretations of how to use the velocity values and integrate them into the distance calculation.

Contextual Notes

There are unresolved issues regarding the integration process, the inclusion of a constant in the anti-derivative, and the correct interpretation of velocity values in relation to time.

Who May Find This Useful

Students or individuals interested in understanding the relationship between velocity and distance in calculus, particularly in the context of physics problems.

tremain74
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Homework Statement
Distance by speed of a particle
Relevant Equations
I want to see if I am using the right technique for this problem. The speed of a particle is given by v = 2*t + 5*t^2, where t is in seconds. What distance does it travel while its speed increases from 7 to 99 m/s?
My solution: v = 2*t + 5*t^2. dx/dt = 2*t + 5*t^2. Therefore dx = (2t + 5t^2)dt. After taking the anti derivative my equation is s = t^2/2 + 5/3(t^3).
 
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That's good so far. Your problem says that the speed goes from 7 to 99, but your equation for ##s## is in terms of time, ##t##. So what is your next step?

PS. A little knit-picking: 5/3(t^3) is easily confused as ##\frac {5}{3(t^3)}##. It would be better to write 5(t^3)/3.
Also, an anti derivative includes a constant, ##c##. The constant does not change the value of the definite integral.
 
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At first I was going to use the values 7m /s to 99 m/s as the numbers to plug into the equations to solve for the distance but these values are for velocity and not for seconds.
 
You ignored the 2 in the 2t term.

You can get times from the velocity equation using the given velocities.
 
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