Answer
(a) $v_x(t) = (0.280~m/s)+ (0.0720~m/s^2)~t$
$v_y(t) = (0.0570~m/s^3)~t^2$
(b) The squirrel is 3.31 meters from the initial position.
(c) The magnitude of the velocity is 1.57 m/s and the direction is an angle of $65.9^{\circ}$ above the positive x-axis.
Work Step by Step
(a) $x(t) = (0.280~m/s)~t+ (0.0360~m/s^2)~t^2$
$v_x(t) = \frac{dx}{dt} = (0.280~m/s)+ (0.0720~m/s^2)~t$
$y(t) = (0.0190~m/s^3)~t^3$
$v_y(t) = \frac{dy}{dt} = (0.0570~m/s^3)~t^2$
(b) When t = 0, the squirrel is at the origin.
At $t = 5.00~s$:
$x = (0.280~m/s)(5.00~s)+ (0.0360~m/s^2)(5.00~s)^2$
$x = 2.30~m$
$y = (0.0190~m/s^3)(5.00~s)^3$
$y =2.38 ~m$
We can find the distance $d$ from the origin.
$d = \sqrt{x^2+y^2}$
$d = \sqrt{(2.30~m)^2+(2.38~m)^2}$
$d = 3.31~m$
The squirrel is 3.31 meters from the initial position.
(c) At $t = 5.00~s$:
$v_x = (0.280~m/s)+ (0.0720~m/s^2)(5.00~s)$
$v_x = 0.640~m/s$
$v_y = (0.0570~m/s^3)(5.00~s)^2$
$v_y =1.43 ~m/s$
We can find the magnitude of the velocity.
$v = \sqrt{v_x^2+v_y^2}$
$v = \sqrt{(0.640~m/s)^2+(1.43~m/s)^2}$
$v = 1.57~m/s$
We can find the angle above the positive x-axis.
$tan(\theta) = \frac{1.43}{0.640}$
$\theta = tan^{-1}(\frac{1.43}{0.64}) = 65.9^{\circ}$
The magnitude of the velocity is 1.57 m/s and the direction is an angle of $65.9^{\circ}$ above the positive x-axis.
