An iceboat on a frozen lake has acceleration a(t)=2tı^+(1−4t3)ȷ^a\left(t\right)=2t\operatorname{î+\left(1-\frac{4}{t^3}\right)ĵ}. At t=4st=4s, the iceboat's velocity is v=8ı^+4.125ȷ^v=8\operatorname{î+4.125}\operatorname{ĵ}. What is the iceboat's displacement between t=2st=2s and t=10st=10s?

267ı^+48.8ȷ^⁡⁡267\operatorname{î+48.8\operatorname{ĵ}}

An iceboat on a frozen lake has acceleration a ( t ) = 2 t ı ^ + ( 1 − 4 t 3 ) ȷ ^ a\left(t\right)=2t\operatorname{î+\left(1-\frac{4}{t^3}\right)ĵ} . At t = 4 s t=4s , the iceboat's velocity is v = 8 ı ^ + 4.125 ȷ ^ v=8\operatorname{î+4.125}\operatorname{ĵ} . What is the iceboat's displacement between t = 2 s t=2s and t = 10 s t=10s ?

All right guys, so welcome back. Let's check out this practice problem. We've got an acceleration time function here. We're told that T equals 4. The velocity is some vector, and we're trying to figure out what the particles, or sorry, what the ice boat's displacement is between 2 and 10 seconds. So what's that variable displacement? Well, that's going to be delta R, and we're looking between two points, a T initial and a T. Final. So let's take a look at a roadmap here and figure out what it is exactly that we're going to do. So we're given an acceleration function, but we're asked to find something about the position, specifically the displacement. Now remember, the displacement you get by doing the integral, and that's going to be a definite integral from T initial to Tf of our velocity time function without the integration constant. So here's the deal. I'm given an acceleration function, but in order to find the position, I'm going to need to find what the velocity function is first. So the first thing I need to do is work backwards to get a velocity time function by taking the integral of the acceleration. So that's going to be the first step. We're going to get the velocity time function from the acceleration time function. So we're going to do the indefinite integral of this mess over here. So this is going to be 2 t in the i direction plus, and this is going to be 1 minus 4 overt. Cubed in the J and we add that integration constant. So now we just go ahead and do that integral. So this becomes T2i plus, and then the integral of 1 is T. And what about this guy? What about 4 overt cubed? That's something new. I haven't seen that before yet. So if we rewrite this 4 overt cubed, remember that when you have X, when you have variables in the denominator, when you have 4. Over T cubed, you can rewrite this as 4 times T to the -3. That negative sign means that you can basically flip it to the denominator. So how do we do this integral? We actually can just follow the same exact rules that we've been following all along. So you're going to do 4 T, now you're gonna raise the exponent by 1, that means add 1 to it. And so it's not going to turn into -4, it's actually going to go, because it's a negative number to -2. 2. So it's going to be 4t to the -2 over -2, and then you can simplify this by writing this as 2 times or 2, sorry, divided by t squared, because remember we can always take this negative power and we can flip it to the bottom. So this is actually what this integral becomes. You have to be careful though because you've now just picked up a negative sign. So what happens is you're going to be doing t minus. Negative 2 overt 2. And so these turn into positives. Now you have this integration constant. So this is our velocity time function, but remember we have to solve for our integration constant. So we're going to do that over here. We do that by using the initial conditions. So if we, if we go back up to the problem here, it tells us that at T equals 4, the velocity is equal to this vector. That's our given initial condition. So this means that at V, So V when T is equal to 4 seconds, is going to be 8i plus 4.125 J. And so, if we plug in T equals 4 inside of this equation over here, we're just going to get 16. In the eye direction, right, cause that's 4 squared, and they're gonna do plus, this is gonna be 4 + 4/4 squared. I'm sorry, that's 2, so 2/4 squared. And that's in the J direction. And that's going to equal, sorry, I forgot to add the integration constant. So plus C is equal to 8I plus 4.125 J. So now what happens is this just becomes 16 i plus 4.125. Just use your calculators to plug that in. Plus C is equal to 8i plus 4.125j. This sets up the equation that we need to solve for that's plus c. And now we just subtract the i's and the j's, just subtract the unit vectors. So because we have a 4.125 on both sides, they'll end up canceling. right, because you move it over to the other side. And then when you bring the 16 over to the other side, you're going to get c equals -8 in the i direction. So this is actually our integration constant. This is the plus c. So we're going to move it back inside of this equation here, but because it has an i hat component or a little i there, we can actually kind of just absorb it or merge it back into the i component that we got for the velocity function. So what this ends up turning out to is v oft is just equal to t2 minus 8, right? That's what happens when we bring this over. Plus 2 + 2 overt 2. J. And so this is our velocity time function. Now, we can actually take the displacement by doing the definite integral of this VT function. So that's really what we've been trying to work to all along. So the second step here is we're going to take the delta R by taking the integral from 2 to 10 seconds of this function over here. So this is T2 minus 8 in the I + T + 2 over T2 in the J. And so, when you actually do this integral here, you're going to get T cubed over 3 minus 8T. That's gonna be in the I direction, plus, this is gonna be T2 over 2. And then what happens is when you do the integral of 2 over t2, you're going to have to use that same rule again, the integral of 2 t to the -2. That's how we rewrite that, is we're going to do 2 t and then we're going to raise the exponent by 1 and divide by that new exponent. So what happens is this actually just becomes -2 overt. And so this is going to be t2 over 2 minus 2 over t, and then we don't have to add our integration constants. Now we basically just evaluate this thing from t equals 2 to t equals 10 seconds, and that's going to be our delta R. So we always do final minus initial. This is going to be the term that we use first, and honestly guys, the rest of this is just going to be plugging and chugging a bunch of numbers into your calculators. So you use Calculators a lot. So you've got 10 cubed over 3 minus 8 times 10. That's the i hat plus, and then we've got 10 squad over 2 minus 2/10. That's in the J hats. That's the first term, that's the final. And now we're going to do minus the initial term. So this is going to be 2 cubed over 3. -8 times 2, that's in the I. Plus, now we've got 2 squared over 2 minus 2/2 in the J, and then you're just going to plug a whole bunch of stuff in. I, I suggest heavily that you plug in each one of these separately and write them down, and then you're just going to subtract the I components from the I components. So I'll highlight those, so you subtract the I's from the I's and then the J's from the J's. And what you should get, once you plug all the stuff in, is you're just going to get delta R equals. 267 I plus 48.8 J. And so this is our displacement between those two times. So it's a little tedious, but if you follow the steps and do these integrals, then you'll get the right answer, right? So the, the answer choice that we're looking for is answer choice C. Let me know if you guys have any questions. Drop me a comment if any of that was confusing. All right guys.

An iceboat on a frozen lake has acceleration a(t)=2tı^+(1−4t3)ȷ^⁡a\left(t\right)=2t\operatorname{î+\left(1-\frac{4}{$$t^3$$}\right)ĵ}. At t=4st=4s, the iceboat's velocity is v=8ı^+4.125⁡ȷ^⁡v=8\operatorname{î+4.125}\operatorname{ĵ}. What is the iceboat's displacement between t=2st=2s and t=10st=10s?

267ı^+48.8ȷ^⁡⁡267\operatorname{î+48.8\operatorname{ĵ}}

330ı^+46ȷ^⁡⁡330\operatorname{î+46\operatorname{ĵ}}

253ı^+49.8ȷ^⁡⁡253\operatorname{î+49.8\operatorname{ĵ}}

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An iceboat on a frozen lake has acceleration $a (t) = 2 t \hat{ı}+(1 - \frac{4}{t^{3}})\hat{ȷ}$ . At $t equals 4 s$ , the iceboat's velocity is $v = 8 \hat{ı}+4.125 \hat{ȷ}$ . What is the iceboat's displacement between $t equals 2 s$ and $t equals 10 s$ ?

$330 dotless i hat plus 46 dotless j hat of$

$253 dotless i hat plus 49.8 dotless j hat of$

$267 dotless i hat plus 48.8 dotless j hat of$

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Problema precedenteProssimo problema

Guida verificata passo dopo passo

Identify the given acceleration vector as a function of time: $\mathbf{a}(t) = 2t \hat{\imath} + \left(1 - \frac{4}{t^3}\right) \hat{\jmath}$.

Recognize that velocity is the integral of acceleration with respect to time. To find the velocity function $\mathbf{v}(t)$, integrate each component of $\mathbf{a}(t)$ separately:

\[v_x(t) = \int 2t \, dt, \quad v_y(t) = \int \left(1 - \frac{4}{t^3}\right) dt.\]

Use the initial condition given at $t=4\,s$, where $\mathbf{v}(4) = 8 \hat{\imath} + 4.125 \hat{\jmath}$, to solve for the constants of integration in both $v_x(t)$ and $v_y(t)$.

Once $\mathbf{v}(t)$ is known, find the displacement $\Delta \mathbf{r}$ between $t=2\,s$ and $t=10\,s$ by integrating the velocity function over this interval:

\[\Delta \mathbf{r} = \int_{2}^{10} \mathbf{v}(t) \, dt.\]

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A toy airplane is flying in an $x/y$ plane parallel to the ground with a velocity $v (t) = (4 t - 0.015 t^{3}) \hat{ı} + (1.8 + 0.5 t) \hat{ȷ}$ . What is the magnitude of the airplane's acceleration at $t equals 30 s$ ?