Find the distance given a function for v(t)

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SUMMARY

The discussion centers on calculating distance using the velocity function v(t) = 2t + 5t². The user correctly derives the equation for distance as s(t) = t²/2 + (5/3)t³, but highlights the need to convert velocity values from 7 m/s to 99 m/s into corresponding time values for accurate distance calculation. Additionally, the importance of including a constant of integration in the anti-derivative is emphasized, as it does not affect the definite integral's value. The user also points out potential confusion in notation regarding the term 5/3(t³).

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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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