(II) At t = 0, a particle starts from rest at 𝓍 = 0 , y ...

Question

(II) At t = 0, a particle starts from rest at 𝓍 = 0 , y = 0 and moves in the xy plane with an acceleration a (→ above a) = (4.0 î + 3.0 ĵ ) m/s². Determine

(a) the 𝓍 and y components of velocity,

Accepted Answer

Hey, everyone in this problem, we're told that a car experiences an acceleration a of 2.5 I hat plus 1.8 J hat meters per second squared in the xy plane starting from a stationary position at T equals zero. We're asked to find both the horizontal and vertical components of velocity. We're given four answer choices and, and we're gonna come back to these answer choices as we work through the problem. So let's start with the horizontal component. And we want to think about the relationship between the velocity which we're trying to find and the acceleration that we're given, right? And recall that the acceleration is the derivative of velocity. OK? Or if we put this the other way around the velocity is the integral of the acceleration. OK. So our change in velocity delta VX in the horizontal direction, it's just gonna be the integral from our initial time zero to the time we're interested in which we'll call T of our horizontal acceleration A XDT. Now this change in velocity delta VX that's gonna be the velocity at time T that we're interested in minus the initial velocity, the X at zero and this is gonna be equal to the integral from zero to T of 2.5 m per second squared DT. And that 2.5 m per second squared comes from the X component that I hot component of our vector A, right? So we're gonna keep moving along here now, VXFT and that horizontal component of the velocity, that's what we're trying to find the velocity at times zero. We're told that we start from a stationary position at times zero. So the velocity is going to be zero. So we have VX at T minus zero, it's just going to be equal to and we have a constant, 2.5 m per second squared is a constant. When we integrate that we're gonna end up multiplying it by T. OK. So we end up with 2.5 meters per second squared multiplied by T. And we have to evaluate that at our endpoints. So from zero to T, when we do that, we get that VX at T is just equal to 2.5 m per second squared multiplied by T because when we evaluate that at zero, we just get zero. So this is the horizontal velocity. Now, let's take a look at the answer choices. The horizontal velocity for option A, we have that it's 2.5 m per second. A constant which is not what we found. Option B we have 2.5 m per second squared multiplied by T squared again, not what we found. Option C 2.5 m per second squared multiplied by T. So that looks good. That looks like what we found. And option D is 2.5 m per second divided by the time T OK. So C is looking like it's gonna be the correct answer. It has the correct horizontal velocity. Let's do the Verti or the, yeah, the vertical velocity just to double check and make sure that we're on the right track and make sure we know how to do this problem. So Delta Vy, we're gonna follow the same hot. OK. So the same equation, the change in velocity is going to be our integral. We're gonna integrate from zero to T 1.8 m per second squared DT. OK. So in this case, that 1.8 m per second squared comes from the J hat component or the vertical component of our acceleration. OK. On the left hand side, this is just gonna be Vyft that vertical velocity we're looking for minus zero because the initial velocity is zero. When we integrate, we get 1.8 meters per second squared multiplied by the time T. And again, we're evaluating from zero to T when we evaluate from zero to T at zero, this gives us zero. OK. So all we end up with Vyft is going to be equal to 1.8 meters per second squared multiplied by the time OK. So the exact same process is on the left hand side and this is going to be our vertical velocity. Comparing to the answer choice, we thought that it was going to be C and that looks indeed like it is the correct answer. So we have option C is correct. A horizontal velocity of 2.5 m per second squared multiplied by T and a vertical velocity of 1.8 m per second squared multiplied by T. Thanks everyone for watching. I hope this video helped see you in the next one.

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(II) At t = 0, a particle starts from rest at 𝓍 = 0 , y = 0 and moves in the xy plane with an acceleration a (→ above a) = (4.0 î + 3.0 ĵ ) m/s². Determine
(a) the 𝓍 and y components of velocity,

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