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Problem 15 Medium Difficulty

A turtle crawls along a straight line, which we will call the $x$-axis with the positive direction to the right. The equation for the turtle's position as a function of time is $x(t) =$ 50.0 cm + (2.00 cm/s)$t -$ (0.0625 cm/s$^2)t^2$. (a) Find the turtle's initial velocity, initial position, and initial acceleration. (b) At what time $t$ is the velocity of the turtle zero? (c) How long after starting does it take the turtle to return to its starting point? (d) At what times $t$ is the turtle a distance of 10.0 cm from its starting point? What is the velocity (magnitude and direction) of the turtle at each of those times? (e) Sketch graphs of $x$ versus $t, v_x$ versus $t$, and $a_x$ versus $t$, for the time interval $t =$ 0 to $t =$ 40 s.

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University Physics with Modern Physics

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Kyle G.

June 16, 2021

Find the turtle’s initial velocity, initial position, and initial acceleration.

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June 16, 2021

Find the turtle’s initial velocity, initial position, and initial acceleration.

Muskan D.

May 11, 2021

A turtle crawls along a straight line, which we will call the x-axis with positive direction to right. The equation for the turtle’s position as a function of time is (a) Find the turtle’s initial velocity, initial position and initial acceleration. (

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Video Transcript

this problem has many parts and so I try and do them as quickly as they can, but also as complete as I can. So we start off with ex of tea being equal to fifty plus two tea minus zero point zero six to five t squared for the duration has promised be enough. I'm going to remove the extra digits, the extra zeros and then also I'm going to remove the units. And remember that distance is measured in centimeters and time is in seconds. Now, from this, we can instantly get the initial position by plugging in. Tea is equal to zero. So when we do that, we get that the initial position is fifty seven years. To get the ft, we simply take the derivative of X ft. With respect the time. It's only do that we get to because the power here is one. And we bring down this too, using the power law. And whenever you multiply this two, by this point zero sixty five, we it zero point one two five t. And so this tells us that fee of zero is equal to two point zero zero centimeters per second and now I want to get the acceleration respected. Time we take the drone of e ft. And so we just get negative point one two five. So the initial acceleration is also negative point one two five because this expression is not the penalty. And so this is the answer to party now. Part B. Ask for the time until the velocity zero. So what we do is we take our fifty from party when we played in Tia's er be a sequel is here only saw for tea and so is equal to zero implies that point one two five t is equal to two. This implies that tea is equal to sixteen seconds. And so that's the time it will take for the turtle too. Get of losses zero. And then after this time, he starts to go back the way you came to the left. But we go ahead and start. Proceed in part. See, we're solving for the time it takes for the turtle to get back to where he started and where he started was at fifty centimetres because on party we saw for the initial distance was fifty seven years. So I'm going to go back to this equation. I'm going to plug in fifty seven years for this. You'LL notice. Why would I do that? I can cancel the fifties and then I can move this term over since it's already negative and make it positive whenever I do that I get. And just to remind you, this Parsi zero point zero sixty five t squared is equal to two tea. At this point, I can cancel a tee and divide both sides by point zero sixty five. Why do that? I get that he is equal to thirty two seconds, and so that should be the answer to part see, so onward to party so part. Do you? We're going to have to use the quadratic formula because now we're given that, except tea is ten centimeters away from the initial starting point. And since the initial starting point is fifty except tea is now sixty and so we want to know the time when this is true. Plugging this into original equation for except e we get that ten equals thirty minus zero point zero sixty five t squared. This is the form for use of quadratic formula. I'm just going to put it in that nice form standard form. I needed a zero point zero six to five t squared and we plus two tea we got minus ten is equal to zero. So now we're going to use the contract formula this term, years or a this term here's or being the stream. Here's or see. And so, by planning project for him, we know that T is equal to negative. Be so I get negative to plus or minus the square root of B squared will be is too. So be squares for minus four. A is negative. Don't forget the negative six to five. And then way also need C in there and see is negative in a closed Beirut and then all over two A. Yes, I'd be two times negative. Zero point zero sixty five. At this point, it's calculator, and if you plug this into your calculator, you get two solutions. Because this is an original quadratic equation. The two solutions are that is equal to six point two one seconds and he is equal to twenty five point eight seconds and you will see there's two times where the turtle is ten seven years away from its starting point. And these were the two times we also need the velocities, magnitude and direction during these two times. And so the way we get that as we just plug him back into the velocity equation v a t. And so the six point two one seconds I'm not going to do all the calculations out. I've already done them on my calculator, but what you get when every plebiscite is one point two two four centimeters per second. So this is the magnitude the direction we get from the sign of this. And since this is positive, we know that this is to the right. Now I have to do the other one. So the year of twenty five point eight and again, this is a calculator problem at this point, and we get negative one point two two four sailors for second. You'LL notice that they're the same man in two. But there opposite signs. This sign implies that now the turtles moving to the right and I believe that is the end of party. There's one more part to this, and that is party in party. We have to Graff exit E V A T and f t. And this could be kind of complicated because the equations air bit complicated. The way I like to do it is by plugging in some points and then just using knowledge of the shapes of quadratic ce and Lanier's in order to really get the overall shape of the same. So I'm going to start with X versus T. And so this could be t and this complex. So we already calculated in party except zero is equal to fifty. So here it is. You're a point. We know that it's going to be at about fifty. If this suspect here and then we also can calculate exit forty, which is supposed to be the end of our domain for the tea here. So ex affording If you go and put this in the calculator, you'LL find that it is equal to thirty. And so if we call this forty, then we go up to roughly here they put a data point. At this point, I'm just going to use the knowledge that this follows a quadratic. If you were called here, this equation is a quadratic and this term here in front of T squared is a negative meaning. It's con cave down if you recall from your calculus class, and so I know that this is going to go up and that's going to come back down. This also agrees with our previous results that at six seconds it's moving to the right, but then at a later time is moving to the left. And so we would expect our ex to increase. And then, as he as the turtle turns around and moves back, it's going to go down again. That's just exit versus tea. We also need V versus T. So if that's team and that is being now, we can do the same thing via zero is, too. It's really two centimeters per second, but I'm ignoring the units, and then V of forty is equal to negative three. And so let me just plot those. We can call that, too, and then out here forty, I'm going to extend my will, access down, and we're going to go a little farther down and so we can call that negative three. Now what happens in between? So we know from party that V is linear. And so what must happen in between is just a straight line between these two. And so it looks like that now, lastly, we havea versus T, which is actually the easiest one. So again, or X axis is T and R Y Axis is going to be a this time. No, we go back to party again and we see that safety is constant. It's not depending on time at all, and it's a constant negative point one two five. What that looks like is just a horizontal line. Atnegative point one two five. And so it looks like that it just extends on forever both ways. But since we wanted from T is equal, zero to tea is equal forty. We're really interested in just that park here. But it's the constant line that negative point one, two five. And so I believe that is all the parts. This problem. I hope I didn't miss anything. Have a good day

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An astronaut has left the International Space Sta…