Find how far a particle travels in 9 seconds

[jesuslovesu]

(from rest)

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

<br /> \int_0 ^{0.0025} |v(t)|dt<br />
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.

 

[moo5003]

What are you plugging in for V not and the starting position X not. Otherwise your integration seems correct.

 

[jesuslovesu]

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

 

[moo5003]

Well... if you solve it that way your getting miles. I suggest converting it to feet.

 

[jesuslovesu]

It turns out to be .0178 ft with the above integral

I also tried the upper limit as 9, which returns 240 ft

 

[moo5003]

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.

 

[jesuslovesu]

Yep, 3.375x10^-6 miles x 5280 ft/mi = .0178 ft

 

[moo5003]

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).

 

[moo5003]

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.

 

[jesuslovesu]

got it thanks