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ARCALcULUSAM ^particle moves along the x-axis with its position given by x(t) = 42t2 St' 100 for tImes t 2 0 Is the speed of this particle increasing or decrea...

Question

ARCALcULUSAM ^particle moves along the x-axis with its position given by x(t) = 42t2 St' 100 for tImes t 2 0 Is the speed of this particle increasing or decreasing at time 17 lustily your answer and show the computations that lead to your conclusion:

ARCALcULUSAM ^particle moves along the x-axis with its position given by x(t) = 42t2 St' 100 for tImes t 2 0 Is the speed of this particle increasing or decreasing at time 17 lustily your answer and show the computations that lead to your conclusion:



Answers

The graph in the st-plane of a position function $s(t)$ of a particle moving rectilinearly is given in FIGURE 4.1.6. Complete the accompanying table by stating whether $v(t)$ and $a(t)$ are positive, negative, or zero. Give the time intervals on which the particle is slowing down and the intervals on which it is speeding up.

In this question, the position off particle function is given us a softy if equal toe 100 by T Square plus 12. Now the velocity function in terms off peak and rewritten as differentiation off as that is, as Dashti respect Toe De. So this gives us minus 200. Be by T Square plus 50 square plus 12 whole race toe to now we need to find the maximum velocity. So to do that view, Dusty must be equal to zero. So we have you dash off D s minus 200 by D squared plus 12. Whole race to bless 800 d squared by. This will be just be. Actually, it's under D by teeth squared, plus 12 whole race toe three. Now this would be equal to zero. Still, we have three d squared, minus 12 equals zero. So did excuse me the time as two seconds. So when the time is two seconds, the velocity off the particle list maximum. So velocity at time to would be minus 200 divided by two squared plus 12 the whole squared. So it would be minus 200 by four plus Well, that would be 16 whole squared. Now this gives me approximately minus one point six feet. But thickened now, when the brillantes odious maximum, We need to find the direction off the particle full. We just substitute toe in the particles function. So we have 100 derided by to square plus 12 which gives me 100 by 16 which is a positive value. So by this I know that the particle is moving to the right, so particles well, toe right.

Here's a nice problem that helps us reason conceptually about what we know about position, velocity and then the rate of change of velocity. So in both graphs were given the position vertically versus time, and were asked to consider what is the velocity doing in particular? Is the velocity increasing or decreasing? So recall that the velocity is the rate of change of position with respect to time. So let's think about what is the derivative of these graphs doing and in particular what is the derivative of that derivative doing? Right. So here we notice the derivatives, which is the slopes of these tangent lines. We see that all the slopes themselves are positive. So that means the velocity would be positive, moreover, notice that the tangent line slopes are getting bigger each time, right? The tangent line slopes are increasing. So here not only is the velocity positive, but the velocity would be increasing as well. Again, that's because the slopes of the tangent lines while positive are becoming more positive. By contrast, Look at the next graph, there are tangent line slopes are still positive, right? They're just getting less and less and less positive. It looks like it's even becoming horizontal, which would have a slip of zero. So here, if we think of the velocity as the derivative of velocity would be decreasing, that's because the rate of change of the slopes is becoming less, so it becomes less positive, thus the velocity decreasing in that picture.

The velocity of the particle er time. Less than five second is provided in the problemas VT 3.2 perimeter per second, so Bt 16.0 minus 1.53 minus 5.0 meter per second. She makes ability is a 4 to 7 meat of a second. The acceleration of the particle is given with the acceleration expression given about the acceleration off. The particle at time, less than for point of five seconds, is given by native Okay. E t is D by D T 3.30 23.2 m bar, second square The acceleration off the particle between times Please bless the greater than five seconds and lesson in seven seconds is given by 30 Stay by ditty, 16.1 point five day minus 5.0 Because my next 1.5 Metro bar said in square, the acceleration of the particle after time take is 11 is given by Ah a d babe, i d 87.0 m par second it is zero me to buy us again square. The position of the particle is given by the expression of the position function from the velocity. The position of the particle for time now he is less than five seconds is given by Ex D is integration 3.2 m part seconded D plus C one, 23.2, please. Whereby to last seven Substitute zero second, 40 and little m four x in the very cool expression who later the rope is a goal today point to little s Square by two save One is say one is a call to zero. Yeah, eso x'd is 3.2 t square by to strike it as one. Now, the position of the particle take called twenties or to to second is calculated by the equation one to his eggs. Two seconds is 3.22 point zero seconds square by two, which makes it 6.4 minutes. Uh huh. Trump, Now the position of the particle between time RT is greater than five and 11 was given by Ex be his integration 1.5 to minus 5.0 80 plus the toe My 16.0 T minus 1.5 square by two plus 7.5 day plus Saito. Yeah, yeah. Hi. A take 4 to 5 seconds. It will be x five point little second. Mm. 3.25 point zero second hole square by two. Which makes it 40.0 minutes from the ever be calculation. Baloo off except five can be related as 40.0 on the corpus 16.0, but 0.0 s minus 1.5510 s Well, square by toe. Last 7.5 My body zero Soto My makes a Saito is a cold. Oh, minus 58.75 Uh huh. Now again, ex Uh, yes, X a d 16.0 t. That's 1.5 k square by two plus seven point five day minus 5. 58.75 has taken us to the version of the particularity. Equal the seventies. Calculated by excellent 0.8 16.7 point little seconds. My nose 1.57 square by two plus 7.5 and 27.0 9. 58.75 Make this extreme nine minutes. The position of the particle at is better. No record of zero ISS X d is a Porto integration seven with nemesis deity plus a 3 to 7.0 T plus three. The constant C is calculated by taking the initial condition that is his initial condition. Table to 11, then 40 40 is the last 40 equal to 11. Then we calculated as 16.0 11 seconds, minus 1.5 11.0 2nd Hold square by two 7.5 11.0 2nd witness. 58.75 It's a nice IT 109 from the above revelation. The low effects at XT at 11.0 2nd is 109 minutes 7.0 11.0 seconds. Last three, Mike said three is 32 so they do em. Yeah, no again X'd is 7.0 t plus 32 AM This is the position for particular tape was to 12. So x 12.0 seconds is 7.0 12.0 plus 42 i just 116 m and the position of the particle AT T to 6.4 30. 70 69.69 AM and that off time 12 with 116 hm

Yeah, for this problem, we are told that a particle moves along the X axis so that its position at any time t greater than or equal to zero is given by X. Of T. Were then asked to find the velocity at the indicated value of T where X of T equals tan inverse of T and T equals two. So the first thing that we can do is the tan inverse or the inverse tan, also known as arc tanne of teeth. Something that is directly differential. The derivative of R kind of thi is one over T squared plus one. So all we need to do now is plug in T equals two. So we have ex prime of two is going to be won over four plus one. So our final result is going to be 1/5.


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