Find how far a particle travels in 9 seconds

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Homework Help Overview

The problem involves determining the distance a particle travels over a period of 9 seconds, given an acceleration function. The acceleration is defined as a = 1 + 3√t mph/sec, and the original poster attempts to find the distance using integration of velocity.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the integration of velocity to find distance and question the units being used. There are attempts to clarify the initial conditions, such as the values for initial velocity and position. Some participants suggest converting units from miles to feet to address discrepancies in the results.

Discussion Status

The discussion is ongoing, with participants exploring different interpretations of the integration process and unit conversions. Some guidance has been offered regarding the need to maintain consistent units throughout the calculations, but no consensus has been reached on the correct approach or final answer.

Contextual Notes

There is mention of the original poster starting from rest, which implies initial conditions of zero for both initial velocity and position. Additionally, there are constraints related to unit conversions between miles and feet, as well as the integration of acceleration in different time units.

jesuslovesu
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I need to find how far a particle travels in 9 seconds
The only given is acceleration but I can find the antiderivative for velocity and position.a = 1 + 3 sqrt(t) mph/sec
v = t + 2t^(3/2) + vo mph
s = 1/2t^2 + 4/5*t^(5/2) + vo*t + so miles

9sec == .0025 hour

[tex] \int_0 ^{0.0025} |v(t)|dt[/tex]
When I take the integral of velocity to find distance traveled, I don't get the correct answer. (should be 344 feet) I would assume that I'm doing something wrong with the units but I can't figure it out.
 
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What are you plugging in for V not and the starting position X not. Otherwise your integration seems correct.
 
I'm just putting in 0 for both vo and so
because from the info that I'm given it says it's from rest
 
Well... if you solve it that way your getting miles. I suggest converting it to feet.
 
It turns out to be .0178 ft with the above integral

I also tried the upper limit as 9, which returns 240 ft
 
Plug in .0025 hours for your equation: .5T^2 + (4/5)T^(5/2)

You will then get miles since you integrated miles/hour.

Since you want your answer in feet you need to convert your answer by multiplying how many feet are in a mile.
 
Yep, 3.375x10^-6 miles x 5280 ft/mi = .0178 ft
 
I'm getting .06525 miles when I plug in numbers.

Although I'm plugging in seconds and then dividng by 3600 (Making mph into mps).
 
I'm getting .06525 miles when I plug in numbers.

Although I'm plugging in seconds and then dividng by 3600 (Making mph into mps).

I tried your way and it does seem to be giving the wrong answer. I believe this is because you integrate acceleration in terms of seconds and then velocity in terms of hours. I believe you have to integrate in terms of one unit and thus divide the velocity equation by 3600. Giving you the same equation over 3600 and you plug in seconds.
 
  • #10
got it thanks
 

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