Answer
(a) $a_x(t) = -(0.0360~m/s^3)~t$
$a_y(t) = (0.550~m/s^2)$
(b) The magnitude of the velocity is 7.47 m/s and the direction is an angle of $59.0^{\circ}$ above the positive x-axis.
(c) The magnitude of the acceleration is $0.621~m/s^2$ and the direction is an angle of $62.4^{\circ}$ above the negative x-axis.
Work Step by Step
(a) $v_x(t) = (5.00~m/s) - (0.0180~m/s^3)~t^2$
$a_x(t) = \frac{dv_x}{dt} = -(0.0360~m/s^3)~t$
$v_y(t) = (2.00~m/s)+(0.550~m/s^2)~t$
$a_y(t) = \frac{dv_y}{dt} = (0.550~m/s^2)$
(b) At $t = 8.00~s$:
$v_x = (5.00~m/s) - (0.0180~m/s^3)~(8.00~s)^2$
$v_x = 3.85~m/s$
$v_y = (2.00~m/s)+(0.550~m/s^2)(8.00~s)$
$v_y =6.40~m/s$
We can find the magnitude of the velocity.
$v = \sqrt{v_x^2+v_y^2}$
$v = \sqrt{(3.85~m/s)^2+(6.40~m/s)^2}$
$v = 7.47~m/s$
We can find the angle above the positive x-axis.
$tan(\theta) = \frac{6.40}{3.85}$
$\theta = tan^{-1}(\frac{6.40}{3.85}) = 59.0^{\circ}$
The magnitude of the velocity is 7.47 m/s and the direction is an angle of $59.0^{\circ}$ above the positive x-axis.
(c) At $t = 8.00~s$:
$a_x = -(0.0360~m/s^3)(8.00~s)$
$a_x = -0.288~m/s^2$
$a_y = (0.550~m/s^2)$
We can find the magnitude of the acceleration.
$a = \sqrt{a_x^2+a_y^2}$
$a = \sqrt{(-0.288~m/s^2)^2+(0.550~m/s^2)^2}$
$a = 0.621~m/s^2$
We can find the angle above the negative x-axis.
$tan(\theta) = \frac{0.550}{0.288}$
$\theta = tan^{-1}(\frac{0.550}{0.288}) = 62.4^{\circ}$
The magnitude of the acceleration is $0.621~m/s^2$ and the direction is an angle of $62.4^{\circ}$ above the negative x-axis.
