- #1

- 3

- 0

Work so far:

Integrated to get these:

Vx(t) = -2t + (3/2)t^2

x(t) = -t^2 + (1/2)t^3

I am just stuck on what to do. I dont need it worked out as much as I just need a push in the right direction. Thanks.

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- Thread starter Brett
- Start date

- #1

- 3

- 0

Work so far:

Integrated to get these:

Vx(t) = -2t + (3/2)t^2

x(t) = -t^2 + (1/2)t^3

I am just stuck on what to do. I dont need it worked out as much as I just need a push in the right direction. Thanks.

- #2

- 15

- 0

- #3

- 3

- 0

PureEnergy said:

So then they are:

Vx(t) = Vi + [-2t + (3/2)t^2]

x(t) = Xi + Vi(t) + [-t^2 + (1/2)t^3]

With Vi meaning initial velocity and Xi meaning initial position.

I just solved it, thanks for the correction.

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