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QuestionTbe wslue af > that miciixes %, where % = 32 + 87, isimaginaryQuestionTbe position at tirne of & particle ix z(t) = t sin(at) . The velccity c the pa...

Question

QuestionTbe wslue af > that miciixes %, where % = 32 + 87, isimaginaryQuestionTbe position at tirne of & particle ix z(t) = t sin(at) . The velccity c the particle &t timeQuestionlim '+02_2_undefined

Question Tbe wslue af > that miciixes %, where % = 32 + 87, is imaginary Question Tbe position at tirne of & particle ix z(t) = t sin(at) . The velccity c the particle &t time Question lim '+02_2_ undefined



Answers

For the particle in Problem $49,$ is there any time $t>0$ when the particle is (a) at rest and (b) accelerating in the $x$ -direction? If either answer is "yes," find the time(s).

I told that were given a position vector for a particle is a function of time X's sign. Need to the T. Why is he? And she is four minus T cubed and we're told that it flies off of this path antique was one. And now we want to know where it is. After it's after one second sort equals two. After one second that's that's flat flown off the path and then proceeded along along a straight line along the tangent. So what we can do is take the velocity, find the velocity by taking the time to do this. So we get either the T co sign out of the T one and then minus three T squared and then at one this is where we are And then this is the velocity at one. So now what we want to do is the position at two Is simply the where we were at one plus the change in time. And then I was when I was going to have a constant velocity after we leave the path. And so in that velocity is the velocity that it was that equals one, And Delta T. Here is simply one time unit. So we have simply this plus this. So we have either the co sign, E times co sign of the plus sign of E I'm two and 0 and numerically that's -2, So we're we're in the xy plane at that point, so we're passing through the X. Y plane.

Yes. Were asked to. Um I think that we have a particle follow in this space curve accident because either the T. Y. Is either minus T. N. Z equals cosine of T. And then at Teak was one. it flies off of this curve and proceeds in a straight line and we want to figure out where the particle is at. T. Was too. So we can first figure out the velocity, so that in general the velocity of the X. Is either the T. B, Y is minus E to the minus T and V. C. Is minus sine of T. So at Teak was one, you have the position is E. One over E and co sign of one. The velocity is E -1 over E- Sign of one. And so the position at, well I shouldn't I think shouldn't be too should be should be T. It should be just to position that too. And after it's flowing off the path is where it was on the path and then the change in time and then the blood in the direction that the velocity was when it flew off the path. Delta T. In this case is one. So we get that where it is now is to E zero. Co signer one minus sine of one and numerically that's 5.440 and minus 0.301 Yeah, that's where it is. After 21 2nd after it's flown off the path, which is followed for one second.

I told that we have a particle or a body that moves along this path here as a function of time. So access T squared. Why is T cubed minus 14. And we're told that paktika was too the part of the body leaves the path no longer follows the path whatever it is. So here's our path and at some point it leaves path and it just flies off and so then it flies off in a straight line. Mhm. And we want to know where it is when T equals three. So well we can just figure out where it is routine was too. And that's that. It's on the X axis. Four units away from the origins of 400 Then you can find what the velocity is and that is just the derivative of this or to T. And he asked us to T. V. Y. Is three T squared minus four and disease of chris hero. So at that time it leaves the velocity is for 80. So it's a position where it is so you know it's left now it's flat moving along the straight line. And the position where it is now is where it was at two. Um Plus the change in time time's the velocity to the velocity is constant now because there's nothing there's nothing constraining the part. So we just delta T. Is one. So this just we're now at 880 is where the particles should be at three.

The position of a particle is a function of time is given by this expression here. So x equals four. Um X equals four t. either the T. Um Why it was 16 T. to the 4th and Z was cosine of T. And we're told that T equals zero and it flies off the path and then we want to know where it is at once. So we have this path and somewhere at T. Zero flies off and then it proceeds along a straight line in the direction of whatever the velocity was at that point in the tangent direction. So we can figure out what the velocity is. And that's just for either the T. 2014 Cube and- Sine of T. In the X. Y. And Z directions. So at T. equals zero. We have the position where at is 401. And at that That time he was zero, the velocity is 400. So where we are equals one, is where we were at um T. T equals zero plus the change in time. In this case it's one times of velocity where we left left the curve. And so we just need to add these two. And so we get 801. That's our position after one second after we've flown off the curve which we basically flew off initially, but we had some initial velocity.


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