Answer to Question #239804 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

The displacement along x axis is given as follows:

$d_x = v_{x0}t+\dfrac{a_xt^2}{2}$

where $v_{x0} = 8m/s$ is the initial velocity along the x-axis, and $a_x = 4m/s^2$ is the acceleration along the x-axis. Thus, for $t = 3.5s$, obtain:

$d_x = 8m/s\cdot 3.5s+\dfrac{4m/s^2\cdot (3.5s)^2}{2} \approx 53m$

Similarly for y-axis:

$d_y = v_{y0}t+\dfrac{a_yt^2}{2}\\ d_y = 12m/s\cdot 3.5s-\dfrac{2m/s^2\cdot (3.5s)^2}{2}\approx 30m$

Thus, the displacement vector is:

$\mathbf{d} = (d_x, d_y) = (53, 30)\space m$

Answer. $\mathbf{d} =(53, 30)\space m$.