Answer
(a) $A = 0~m$
$B = 2.00~m/s^2$
$C = 50.0~m$
$D = 0.500~m/s^3$
(b) $\vec{v} = (0\hat{i}+ 0\hat{j})~m/s$
$\vec{a} = (4.00~\hat{i}+ 0\hat{j})~m/s^2$
(c) $v_x = 40.0~m/s$
$v_y = 150~m/s$
$v = 155~m/s$
(d) $\vec{x} = (200\hat{i}+550\hat{j})~m$
Work Step by Step
(a) $x(t)= A+Bt^2$
$v_x(t) = \frac{dx}{dt} = 2Bt$
$a_x(t) = \frac{dv_x}{dt} = 2B$
$y(t)= C+Dt^3$
$v_y(t) = \frac{dy}{dt} = 3Dt^2$
$a_y(t) = \frac{dv_y}{dt} = 6Dt$
When t = 0:
$(x,y) = (0~m,50.0~m)$
$A+Bt^2 = A+B(0~s)^2 = 0~m$
$A = 0~m$
$C+Dt^3 = C+D(0~s)^3 = 50.0~m$
$C = 50.0~m$
When t = 1.00 s:
$a_x = 4.00~m/s^2 = 2B$
$B = 2.00~m/s^2$
$a_y = 3.00~m/s^2 = 6D(1.00~s)$
$D = 0.500~m/s^3$
(b) $\vec{v} = 2B(0~s)\hat{i}+ 3D(0~s)^2\hat{j}$
$\vec{v} = (0\hat{i}+ 0\hat{j})~m/s$
$\vec{a} = 2B\hat{i}+ 6D(0~s)\hat{j}$
$\vec{a} = (4.00~\hat{i}+ 0\hat{j})~m/s^2$
(c) At t = 10.0 s:
$v_x = 2Bt = (2)(2.00~m/s^2)(10.0~s)$
$v_x = 40.0~m/s$
$v_y = 3Dt^2 = (3)(0.500~m/s^3)(10.0~s)^2$
$v_y = 150~m/s$
We can find the speed $v$ of the rocket.
$v = \sqrt{(40.0~m/s)^2+(150~m/s)^2}$
$v = 155~m/s$
(d) At t = 10.0 s:
$\vec{x} = ((A+Bt^2)\hat{i}+(C+Dt^3)\hat{j})~m$
$\vec{x} = ((0+(2.00~m/s^2)(10.0~s)^2)\hat{i}+((50.0~m)+(0.500~m/s^3)(10.0~s)^3)\hat{j})~m$
$\vec{x} = (200\hat{i}+550\hat{j})~m$
