Answer to Question #239802 in Physics for hidee

Question

A cart is travelling over an xy plane with acceleration components ax = 4.0 m/s^2 and ay=-2.0 m/s^2. if the initial velocity along x is 8.0 m/s and along y is 12 m/s. find, the displacement after 3.5 seconds

Answer 1

Let's first find the $x$- and $y$-components of cart's displacement after 3.5 s:

$$d_x=v_{0x}t+\dfrac{1}{2}a_xt^2,$$ $$d_y=v_{0y}t+\dfrac{1}{2}a_yt^2,$$ $$d_x=8.0\ \dfrac{m}{s}\cdot3.5\ s+\dfrac{1}{2}\cdot4.0\ \dfrac{m}{s^2}\cdot(3.5\ s)^2=52.5\ m,$$ $$d_y=12\ \dfrac{m}{s}\cdot3.5\ s+\dfrac{1}{2}\cdot(-2.0\ \dfrac{m}{s^2})\cdot(3.5\ s)^2=29.75\ m.$$

We can find the magnitude of the displacement of the cart from the Pythagorean theorem:

$$d=\sqrt{d_x^2+d_y^2}=\sqrt{(52.5\ m)^2+(29.75\ m)^2}=60.34\ m.$$

We can find the direction of the displacement of the cart from the geometry:

$$\theta=tan^{-1}(\dfrac{d_y}{d_x})=tan^{-1}(\dfrac{29.75\ m}{52.5\ m})=29.5^{\circ}.$$