A particle is moving in x-y plane, its position vector is...

A particle is moving in x-y plane, its position vector is given by \(\vec{r} = [(6 + 3t)\hat{i} + (5 + 2t - 2t^2)\hat{j}]\) m, where t is measured in seconds. a) Find its position vector and velocity vector at \(t = 0\) s (in unit vector notation). b) Find the acceleration of the particle (in unit vector notation). c) How long (in s) will take the particle to reach the maximum y-coordinate?

Transcript

00:01 First, they're giving us a function for the position vector with respect to time.

00:06 And this would be equaling 2 .00 t cubed minus 5 .00 t i .at.

00:20 And then this would be plus 6 .00 minus 7 .00 t to the fourth j hat.

00:34 So we can then say that for part a, the position vector at t equaling 2 .00 seconds, well, we can simply substitute.

00:47 So this would be 2 .00 multiplied by 2 .00 cubed minus 5 .00, multiplied by 2 .00 ihat, plus 6 .00 minus 7 .00.

01:06 Times 2 .00 to the 4th power j hat.

01:11 And so this is giving us 6 .00 i hat minus 106 j hat.

01:25 And the units would of course be meters.

01:30 So this would be our answer for part a.

01:32 For part b then to find the velocity we need to take the derivative.

01:36 So the velocity equation with respect to time would be equaling 6 .0.

01:42 T squared minus 5 .00 i hat minus 28 .0 t cubed j hat and here then the velocity at t equals 2 .00 seconds would be equaling 6 .00 multiplied by 2 .00 seconds quantity squared minus 5 .00 i hat minus 2 .00 i hat minus 28 point zero times 2 .00 seconds quantity cubed j hat and we find that then the velocity at t equaling 2 .00 seconds is going to be equaling 19 .0 i .hat minus 224 j hat and the units of course meters per second this would be your answer for part b for part c we want to find the acceleration so the acceleration function with respect to time, this would be 12 .0 t i hat minus 84 .0 t squared jhat.

03:07 And so solving for the acceleration at t equaling 2 .00 seconds, this would be equalling 24 .0 .0...

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