A bird flies in the xy-plane with a velocity vector given by v = (a - ^2) l̂+γt ĵ, with α= 2.4 m

A bird flies in the xy-plane with a velocity vector given by...

A bird flies in the $xy$-plane with a velocity vector given by $\overrightarrow{v}$ = $(a - \beta$$t^{2})$ $\hat{l}+\gamma t \hat{j}$, with $\alpha = 2.4 m/s$, $\beta = 1.6 m/s^3$, and $\gamma = 4.0 m/s^2$. The positive y-direction is vertically upward. At $t = 0$ the bird is at the origin. (a) Calculate the position and acceleration vectors of the bird as functions of time. (b) What is the bird's altitude (y-coordinate) as it flies over $x = 0$ for the first time after $t = 0$?

Transcript

00:02 So we're told that the bird flies in the xy plane and we're given the velocity vector v, which is equal to alpha minus beta times times squared in the x direction plus gamma times time in the y direction.

00:16 And then it gives us the values for alpha, beta and gamma, which i wrote here.

00:20 In the positive y direction, it is upwards right here vertically.

00:24 So at t equals zero, the bird is at the origin.

00:27 So since t is equal to zero at the origin, that also means that v of x initial is zero and v of y initial is zero here.

00:37 Okay, so for part a, calculate the position and acceleration vectors of the bird as a function of time.

00:43 So let's first find the position vector x, which is going to be the integral of v of x with respect to time.

00:53 So this is going to be the integral of alpha minus beta times time squared with respect to time, which is equal to alpha times time minus one -third beta times time to the third.

01:36 We can box that in as part of our solution for part a.

01:41 Now we need to find the y part of the vector.

01:45 This is going to be the integral of v of y with respect to time.

01:54 V of y is equal to gamma times time.

01:57 So this integral is going to be one half, gamma times time squared.

02:10 We can box that in as the second part.

02:13 Both of those make up the position vector, the x part and the y part.

02:17 Now we need to find the acceleration in the x direction, a of x.

02:24 That's simply the time derivative of the x part of the velocity vector v.

02:32 X so this is going to be equal to there's no time component of alpha so it's just going to be equal to minus the two comes down front two beta times time and then lastly let's rebox that in so we're not blocking out part of the solution lastly we just need to find the acceleration in the y direction a sub y, that's going to be equal to the time derivative of v .y, which is just gamma t...

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