motion question (1 Viewer)

Hermes1 · Apr 6, 2011

if an objects acceleration is given by:

(3t+1)^2 and it is originally at rest 2 metres to the left of the origin. find the displacement after 4 seconds

Drongoski · Apr 6, 2011

arj1 said:
if an objects acceleration is given by:

(3t+1)^2 and it is originally at rest 2 metres to the left of the origin. find the displacement after 4 seconds

$\frac {d^2 x}{dt^2} = (2t+1)^2$

$\therefore \frac {dx}{dt} = \frac {(2t+1)^3}{2 \times 3} + C$

Originally at rest means: when t = 0, $v = \frac {dx}{dt} = 0 \implies C = - \frac {1}{6}$

$\therefore x(t) = \int \frac {(2t+1)^3}{6} - \frac {1}{6} dt = \frac {(2t+1)^4}{2 \times 4 \times 6} - \frac {t}{6} + D$

When t=0, x = - 2 so that D = - 1/48

$\therefore x(4) = \frac {(2 \times 4 +1)^4}{48} - \frac {4}{6} - \frac {1}{48}$

I'll finish the rest if you hire me as your tutor. Otherwise complete the rest yourself.

genius1234567 · Apr 6, 2011

Drongoski said:
I'll finish the rest if you hire me as your tutor. Otherwise complete the rest yourself.

bout time you started being a bit more strict in terms of giving help! lol nice

Drongoski · Apr 6, 2011

Of course that was tongue-in-cheek. I've more or less completed the solution.

hscishard · Apr 6, 2011

I don't blame drongoski for saying "I'll finish the rest if you hire me as your tutor. Otherwise complete the rest yourself."
But if he does become like this guy ^, then yea, he'll be like a recruit

Btw, it's (3t+1)^2

genius1234567 · Apr 6, 2011

hscishard said:
But if he does become like this guy ^, then yea, he'll be like a recruit

have no idea what u are on about

motion question (1 Viewer)

Hermes1

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Drongoski

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genius1234567

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Drongoski

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hscishard

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genius1234567

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