A bird flies in the xy-plane with a position vector given by \vec{r}=(\alpha t-\beta t^3)\hat{i}+\gamma t^2\hat{j}, with \alpha=2.4\ m/s,\beta=1.6\ m/

Dolly Robinson

Dolly Robinson

Answered question

2021-01-24

A bird flies in the xy-plane with a position vector given by r=(αtβt3)i^+γt2j^, with α=2.4 ms,β=1.6 ms3 and γ=4.0 ms2. The positive y-direction is vertically upward. At the bird is at the origin.
Calculate the velocity vector of the bird as a function of time.
Calculate the acceleration vector of the bird as a function oftime.
What is the bird's altitude(y-coordinate) as it flies over x=0 for the first time after ?

Answer & Explanation

Margot Mill

Margot Mill

Skilled2021-01-25Added 106 answers

A bird flies in the xy-plane with a position vector given by r=(αtβt3)i^+γt2j^, with α=2.4 ms,β=1.6 ms3 and γ=4.0 ms2.
r=(2.4t1.6t3)i+4.0t2j
The positive y-direction is vertically upward. At t=0 the bird is at the origin. The velocity vector of the bird as a function of time(v=dr/dt).
v=(2.4t4.8t2)i+8.0tj
The acceleration vector of the bird as a function of time (a=dv/dt)
a=(9.6t)i+8.0j
Jeffrey Jordon

Jeffrey Jordon

Expert2021-10-11Added 2605 answers

The velocity vector is

v(t)=(αβt2)i^+(γt)j^

1. The position vector can be found by integrating the velocity vector.

r(t)=v(t)dt=(αtβt33+C1)i^+(γt22+C2)j^

At t = 0, the bird is at the origin, so the integration constants can be determined.

r(0)=C1i^+C2y^=0

Therefore, C1 and C2 are equal to zero.

r(t)=(αtβt33)i^+(γt22)j^

2. The acceleration vector is the derivative of the velocity vector with respect to time.

a(t)=dv(t)dt=2βti^+γj^

3. We will first find the time when the x-component of the position vector is equal to zero.

αtβt33=0

α=βt33

t=3αβ=2.12s

We will plug in this value into the y-component of the position vector.

y(t)=γt22

y(2.12)=4(2.12)22=9m

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