Respuesta :
Answer:
1. [tex]r(t) = (\alpha t - \frac{\beta t^3}{3})\^i + (\frac{\gamma t^2}{2})\^j[/tex]
2. [tex]a(t) = \frac{dv(t)}{dt} = -2\beta t \^i + \gamma \^j[/tex]
3. [tex]r(t=2.12) = (6~{\rm m})\^j[/tex]
Explanation:
The velocity vector is
[tex]v(t) = (\alpha - \beta t^2)\^i + (\gamma t)\^j[/tex]
1. The position vector can be found by integrating the velocity vector.
[tex]r(t) = \int v(t) dt = (\alpha t - \frac{\beta t^3}{3} + C_1)\^i + (\frac{\gamma t^2}{2} + C_2)\^j[/tex]
At t = 0, the bird is at the origin, so the integration constants can be determined.
[tex]r(0) = C_1 \^i + C_2 \^y = 0[/tex]
Therefore, C1 and C2 are equal to zero.
[tex]r(t) = (\alpha t - \frac{\beta t^3}{3})\^i + (\frac{\gamma t^2}{2})\^j[/tex]
2. The acceleration vector is the derivative of the velocity vector with respect to time.
[tex]a(t) = \frac{dv(t)}{dt} = -2\beta t \^i + \gamma \^j[/tex]
3. We will first find the time when the x-component of the position vector is equal to zero.
[tex]\alpha t - \frac{\beta t^3}{3} = 0\\\alpha = \frac{\beta t^2}{3}\\t = \sqrt{\frac{3\alpha}{\beta}} = 2.12~s[/tex]
We will plug in this value into the y-component of the position vector.
[tex]y(t) = \frac{\gamma t^2}{2}\\y(2.12) = \frac{4(2.12)^2}{2} = 9~m[/tex]
