The coordinates of a bird flying in the xy-plane are given by ...

Question

The coordinates of a bird flying in the xy-plane are given by x(t) = αt and y(t) = 3.0 m − βt², where α = 2.4 m/s and β = 1.2 m/s². Calculate the magnitude and direction of the bird's velocity and acceleration at t = 2.0 s.

Accepted Answer

everyone in this video. We have a problem where we're given a particle's location invariable coordinates. So we have X of T is equal to 0.5 80 m. And y of T is equal to B t squared plus 5.5 m. Okay, we're given values for A and B of 0.63 m per second and minus 0.8 m per second squared respectively. And we're asked to find the magnitude and direction of the particles, acceleration and velocity after half a minute. Okay. Alright. So let's start by writing out the quantities were given. So we're given X of T is equal to 0.5 that's in meters. Okay. And we're told that A is 6.3 m per second. So we have 0.5 Times 6.3 T. And then 6.3 is m per second time is in units of seconds. So we'll be left with the unit of meter and working this out, we are going to have um 3.15 T for X. Okay. Alright. And then same for the Y. We're just going to substitute in the values, we know if Y. T. Is equal to B. T squared plus 5.5, substituting our values, we get minus 0.8 t squared plus 5.5. Okay, so we're talking about velocity, acceleration and the relationship to the position. Let's recall that velocity is going to be the change in position over time. Okay, so we have the position with the velocity in each direction respectively. So the velocity let's say in the X direction is a function of time. It's just going to be the change in the position. X over time. Okay, so we're going to take the derivative of this X. Of T. Okay, so derivative of X. Of T. While it's a linear function. So we take the derivative, the exponents going to reduce by one. We're gonna have just a constant left. So it's going to be 3.15. Similarly in the Y direction, the Y velocity as a function of time is going to be the change in the Y position over time. And taking the derivative of this function, we have a squared term. Okay, so we're gonna have -0.08. We're gonna multiply by the exponent and then the exponents going to reduce by once we're gonna be left with just the T. And the constant, the derivative of constant is just zero. Okay, so that's gonna be minus 0.16 T. Alright, so we found the general equation, I'm going to scroll down so we have more room to work here. We found the general equation for the velocity in both the X and Y direction. Now we need to find it for the specific problem we're being asked. So this problem is asking us to find the velocity after half a minute. Okay, so half a minute is going to be 30 seconds. So our t. Is going to be 30 seconds. Okay, well the velocity of acts or in the X direction at 30 seconds is going to be 3.15. Okay, this is a constant. It doesn't depend on t So no matter what time we get Our velocity in the extraction will still be 3.15. That's going to be meters per second. And in the Y direction, the velocity of y at 30 seconds. Well, that's gonna be minus 0.16 times 30. That's gonna be -4. m/s. Alright, So if you think about this particle moving, we know it's moving to the right with some velocity, Okay, 3.15 meters for a second and we know it's moving down. Okay, because this velocity is negative, so it's going downwards in the Y direction, so it's going to be going down 4.8 m/s. And so the magnitude of the velocity, the velocity is going to be this vector here be okay, the magnitude is going to be just this quantity from the triangle. Okay, we can find that using pythagorean theorem V squared It's equal to 3.15 square. Okay, we're looking for the hypotenuse of the triangle plus 4. squared. So V squared. Working that out, we get 32.96 to 5. And when we take the square root we get V is equal to plus or minus 5.7413 m per second. Okay. And in this case we're looking just for the magnitudes the magnitude of V. We're gonna take the positive route 5. 413 m/s. Okay. Alright. So we found now the magnitude of the velocity, we need to find the direction. Well, we can find this angle here. Okay, we'll call it theta prime. But we have to remember that when we're talking about angles, they're measured from the positive X axis. So if we were to draw our axes like this. Okay, we have X in the positive direction. We have Y going down. So the actual angle we're looking for here, this data. Okay. Well how are they related? Well, we know the entire rotation will be 360 degrees. So data, it's going to be 360 degrees minus data prime. Okay. We're going to write this because we know we can find theta prime. We know that. 10 tangent of theta prime. Okay, because we have theta prime here, we have the opposite side divided by the adjacent side to 4.8 divided by 3.15. Okay, so theta prime is going to be Are 10 4.8 over 3.15. Alright. And that's going to give us 56.725°. Alright. And then go back to our value of data. Data is going to be 360° -56.725°. Okay, so we found this data prime the angle from the X axis down to our vector. But what we want to find is that angle that goes from the positive X axis all the way around. Okay, so we're gonna take 3 60 minus that angle we found and the angle we get is going to be 303.275 degrees. Okay, so that's great. We found the magnitude of the velocity. We found the direction of the velocity. Okay now let's scroll up and see what else we were asked to find. So the magnitude and direction of the acceleration and velocity. Okay, so we've done the velocity Now let's move to the acceleration. Okay. And I'm going to write the quantities we found for V again so we can refer to them. So we found that VX is a function of T is just 3.15 meters per second. And we found that the y is a function of time was -0.1 six T. No, similarly to the relationship between velocity and position. The acceleration is going to be the change over time of the velocity. Okay, so for each component, each direction, the extraction in the y direction, acceleration is going to be changing that corresponding velocity over time. So the acceleration in the X direction as a function of time is going to be D V X. Oddity. Okay, well V X is just a constant. So when you take the derivative we're just going to get zero. So the acceleration in the X direction is going to be zero. The unit here meters per second squared. And in the Y direction we're going to do the same. So the acceleration in the y direction going to be the derivative change in the Y velocity over time. Okay, in this case is going to be -0.16. Okay, so that's a constant. So no matter what time we plug in there, that's the value we're going to get. Okay, so now we want to find the magnitude and direction of the total acceleration. So the magnitude of the acceleration while we only have the acceleration going down. So if we want to draw a little diagram we have this acceleration going down at 0.16 m per second. Okay, and if we draw this in terms of our axis looks like this. Alright, so the magnitude of eight is just gonna be the magnitude minus 0.16 m per second squared. Okay, Which is just going to be 0.16 m per second squared. Right? And then if we want to think of direction again, we're going from the positive X axis, we're going counterclockwise And that's gonna be 270°. Okay, so the angle of the acceleration, I'm just gonna add data v up here to indicate that that's the angle of the velocity. So we don't mix them up. So the angle of the acceleration 270 degrees because it's lying right on that negative y axis. Okay? So here's our acceleration, here's the magnitude and direction. Alright, so let's go back and choose an answer. So again, the magnitude of the acceleration is 0.16 m per second squared. The Direction is 270°. The magnitude of the velocity 5.74 m/s and its angle 303.275°. Okay, so back to our answer choices that is going to correspond to answer B. Thanks for watching. I hope this video helped see you in the next one.

The coordinates of a bird flying in the xy-plane are given by x(t) = αt and y(t) = 3.0 m − βt², where α = 2.4 m/s and β = 1.2 m/s². Calculate the magnitude and direction of the bird's velocity and acceleration at t = 2.0 s.

Key Concepts

Velocity in Two Dimensions

Acceleration in Two Dimensions

Direction of a Vector

Watch next

The coordinates of a bird flying in the xy-plane are given by x(t) = αt and y(t) = 3.0 m − βt2, where α = 2.4 m/s and β = 1.2 m/s2. Calculate the magnitude and direction of the bird's velocity and acceleration at t = 2.0 s.

Key Concepts

Velocity in Two Dimensions

Acceleration in Two Dimensions

Direction of a Vector

Watch next

The coordinates of a bird flying in the xy-plane are given by x(t) = αt and y(t) = 3.0 m − βt², where α = 2.4 m/s and β = 1.2 m/s². Calculate the magnitude and direction of the bird's velocity and acceleration at t = 2.0 s.