Answer
(a) $x(t) = (2.4~m/s) ~t -(\frac{1.6~m/s^3}{3}) ~t^3$
$a_x(t) = (-3.2~m/s^3)~t$
$y(t) = (2.0~m/s^2) ~t^2$
$a_y(t) = 4.0~m/s^2$
(b) The bird's altitude is 9.0 meters as it flies over x = 0 for the first time after t = 0
Work Step by Step
(a) $v_x(t) = \alpha - \beta ~t^2$
$x(t) = x_0+\int_{0}^{t}v_x(t)$
$x(t) = \int_{0}^{t}\alpha - \beta ~t^2$
$x(t) = \alpha ~t -\frac{\beta}{3} ~t^3$
$x(t) = (2.4~m/s) ~t -(\frac{1.6~m/s^3}{3}) ~t^3$
$a_x(t) = \frac{dv_x}{dt} = -2\beta ~t$
$a_x(t) = (-2)(1.6~m/s^3)~t = (-3.2~m/s^3)~t$
$v_y(t) = \gamma ~t$
$y(t) = y_0+\int_{0}^{t}v_y(t)$
$y(t) = \int_{0}^{t}\gamma ~t$
$y(t) = \frac{\gamma}{2} ~t^2$
$y(t) = (2.0~m/s^2) ~t^2$
$a_y(t) = \frac{dv_y}{dt} = \gamma = 4.0~m/s^2$
(b) We can find $t$ when $x(t) = 0$
$x(t) = (2.4~m/s) ~t -(\frac{1.6~m/s^3}{3}) ~t^3 = 0$
$(\frac{1.6~m/s^3}{3}) ~t^2 = 2.4~m/s$
$t^2 = \frac{(3)(2.4~m/s)}{1.6~m/s^3} = 4.5~s^2$
$t = \sqrt{4.5~s^2} = 2.12~s$
At t = 2.12 s:
$y = (2.0~m/s^2) ~(2.12~s)^2 = 9.0~m$
The bird's altitude is 9.0 meters as it flies over x = 0 for the first time after t = 0
