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Numerade Educator



Problem 41 Hard Difficulty

The position of an object with mass $m$ at time $t$ is
$\mathbf{r}(t)=a t t^{2} \mathbf{i}+b t^{3} \mathbf{j}, 0 \leqslant t \leqslant 1$
(a) What is the force acting on the object at time $t ?$
(b) What is the work done by the force during the time
interval 0$\leqslant t \leqslant 1 ?$


a. $<2 m a, 6 m b t>$
b) $2 m a^{2}+\frac{9}{2} m b^{2}$


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Video Transcript

All right, so in this video, we're told position vector. If an object is a 80 swear it I have. Plus B, t g and bestie goes from 01 harassed have determined the floor sacking on the object it Well, we know that forces m times A where a is the second derivative of the position. So we're gonna take the second derivative of our So the first derivative of poor. It's just a derivative off 80 square. Good perspective, T which is to Ratey and derivative of PT Cube respected T, which is sleepy to square. And then now we're gonna take the second derivative of the first. So venting derivative of the first derivatives of the derivatives to 81 respected to use two way and the derivative off three bt squared when perspective to use give six pt. So now we determined Fowler. Excellent. So after sometimes a. So we're just gonna multiply our acceleration factor, Which is to say I have to be t hat. Um, I m so we get to any my at plus two MBT jihad and another condensed waiter. Right? There is an excuse. Giorni Coma six MBT. All right. The second and part B were asked to determine of the work done in moving a particle from teeth was your letter t hose. So we know that work is just the line in general of F d r. And then we can write that in another way. In terms of tea as integral from 0 to 1 of f dot Our prime of TV. All right, we found a F in part A because just two in a comma, six MBT and also in part A. We found our prime of tea, which is just to 80 comma, three bt square. Now we're gonna do performed adult problem. So we get four and a square T plus 18 and the square Todt Cube. All right, so we have a four m. We haven't 18 am What we can pull out as a common factor is it to em? So this is exactly what we do right here. So we're left with integral from zero to a lot of to a square. T plus nine b square T Cuba Bt again. Anything. That's not He's just a constant. So we don't bother with this with this. Just a power rule for Integrys. So the derivative of the integral of teens just he swear, divided by two. So I left with eight square. He squared and they enter growth. T Cube is just to the power for divided by and again are integral goes from 0 to 1. All right, so now we're gonna take we're gonna plug in that integral limits. So we get a squared plus nine divided by four square. This is when we plug 1 40 And if it begins zero perky, we're just going to get minus zero plus zero. So this is zero. So finally, the works? Yeah. The work is equal to two M times A a squared plus nine divided by four square.