Question

A bird flies in the xy-plane with a velocity vector givenby v = (α-βt2)i + γtj with α=2.4m/s,β=1.6m/s3 and γ=4m/s2 . Thepositive y-direction is vertically upward. At t=0 the birdis at the origin.
1. Calculate the position vector of the bird as a function oftime.
2. Calculate the acceleration vector of the bird as a function oftime.
3. What is the bird's altitude (y-coordinate) as it fliesover x=0 for the first time after t=0.

Answer 1

Answer:

1. $r(t) = (\alpha t - \frac{\beta t^3}{3})\^i + (\frac{\gamma t^2}{2})\^j$

2. $a(t) = \frac{dv(t)}{dt} = -2\beta t \^i + \gamma \^j$

3. $r(t=2.12) = (6~{\rm m})\^j$

Explanation:

The velocity vector is

$v(t) = (\alpha - \beta t^2)\^i + (\gamma t)\^j$

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

$r(t) = \int v(t) dt = (\alpha t - \frac{\beta t^3}{3} + C_1)\^i + (\frac{\gamma t^2}{2} + C_2)\^j$

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

$r(0) = C_1 \^i + C_2 \^y = 0$

Therefore, C1 and C2 are equal to zero.

$r(t) = (\alpha t - \frac{\beta t^3}{3})\^i + (\frac{\gamma t^2}{2})\^j$

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

$a(t) = \frac{dv(t)}{dt} = -2\beta t \^i + \gamma \^j$

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

$\alpha t - \frac{\beta t^3}{3} = 0\\\alpha = \frac{\beta t^2}{3}\\t = \sqrt{\frac{3\alpha}{\beta}} = 2.12~s$

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

$y(t) = \frac{\gamma t^2}{2}\\y(2.12) = \frac{4(2.12)^2}{2} = 9~m$