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Old 04-25-2010, 07:49 PM   #1
Phantom Badger
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Physics Question

My mind has just gone blank and I've forgotten how to do this, any help ?

Find the Distance travelled by an object accelerating at 0.4ms^2 over 50 seconds (at the end of the 50 seconds the object is travelling at 20m/s)

Halp my failing brain. I know an object with a constant speed it would be D=ST vut for an accelerating object Im not sure what I should do.

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Old 04-25-2010, 07:55 PM   #2
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v final = v initial + at
20 = v initial + 0.4(50)
v initial = 0 m/s

d = v initial + 0.5at^2
d = 0 + 0.5(0.4 m/s^2)(20)^2
d = 80 m
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Old 04-25-2010, 07:58 PM   #3
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I thought it was

Distance = 1/2Acceleration multiplied by t^2
Not
Initial velocity+1/2a(t^2)

Then again my Physics teacher is crap. He hasnt even taught us this :/
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Old 04-25-2010, 08:21 PM   #4
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Wait just had another look through my exam book and got the equation

d=ut+0.5at^2
which with the Numbers was
D=(0*50)+0.5(0.4*50^2) = 500metres...

And to I also used the other equation the book suggested which was:
d=((u+v)/2)t which again equaled 500.

Triumph...Got a Physics question wrong ?

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Old 04-26-2010, 03:10 AM   #5
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d = vt - 0.5at^2
d = 20 x 50 - 0.5 x 0.4 x 50^2 = 500m
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Old 04-26-2010, 05:04 AM   #6
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Wow that's embarrassing.

I did it two ways and got 80 m and 500 m, the second way was:

(v final)^2 = (v initial)^2 + 2ad
d = [(v final)^2 - (v initial)^2]/2a
d = (20)^2/2(0.4)
d = 500 m

I figured I was wrong, and picked the smallest answer... in my defense I was on the phone and was working on economics, lmao
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Old 04-26-2010, 05:05 AM   #7
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Econ > Physics.
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Old 04-26-2010, 04:51 PM   #8
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Wtf is all this? @_@

-Dies-
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Old 04-26-2010, 05:03 PM   #9
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Originally Posted by Ralath View Post
Econ > Physics.
You can only do that because in Econ, you start off by saying "Let's assume Economics is better than Physics..."
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Old 04-26-2010, 05:06 PM   #10
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Thank gawd we dun study this here. @__@

9th grade is easy. x_x
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