Question

A cart is propelled over a xyplane with acceleration components${\mathbf{a}}_{\mathbf{x}}\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{0}\mathbf{m}\mathbf{/}{\mathbf{s}}^{\mathbf{2}}\mathbf{}\mathbf{and}\mathbf{}{\mathbf{a}}_{\mathbf{y}}\mathbf{=}\mathbf{-}\mathbf{2}\mathbf{.}\mathbf{0}\mathbf{m}\mathbf{/}{\mathbf{s}}^{\mathbf{2}}$.Its initial velocity has components ${\mathbf{v}}_{\mathbf{0}\mathbf{x}}\mathbf{=}\mathbf{8}\mathbf{.}\mathbf{0}\mathbf{m}\mathbf{/}\mathbf{s}\mathbf{.}\mathbf{}\mathbf{And}\mathbf{}{\mathbf{v}}_{\mathbf{0}\mathbf{y}}\mathbf{=}\mathbf{12}\mathbf{m}\mathbf{/}\mathbf{s}\mathbf{.}$In unit-vector notation, what is the velocity of the cart when it reaches its greatest y coordinate?

Short answer

The velocity of the cart when it reaches$\mathrm{y}={\mathrm{y}}_{\max }\mathrm{is}\left(32\mathrm{m}/\mathrm{s}\right)\hat{\mathrm{i}}$

Step-by-step solution

Given information

Acceleration along x and y axes are

$$\begin{gathered} {\mathrm{a}}_{\mathrm{x}}=4.0\mathrm{m}/{\mathrm{s}}^2 \\ {\mathrm{a}}_{\mathrm{y}}=-2.0\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

Initial velocity components along x and y axis are

$$\begin{gathered} {\mathrm{v}}_{0\mathrm{x}}=8.0\mathrm{m}/\mathrm{s} \\ {\mathrm{v}}_{0\mathrm{y}}=12\mathrm{m}/\mathrm{s} \end{gathered}$$

To understand the concept

This problem is based on kinematic equations that describe the motion of an object with constant acceleration. Here, first kinematic equation for vertical motion to find the time for which${\mathbf{v}}_{\mathbf{max}}\mathbf{=}\mathbf{0}\mathbf{m}\mathbf{/}\mathbf{s}$. Using the value obtained for time in the first kinematic equation along x axis, the velocity of the cart when it reaches its greatest y coordinate can be found.

Formula:

The final velocity in kinematic equation can be written as,

${\mathrm{v}}_{\mathrm{f}}={\mathrm{v}}_0+\mathrm{at}$ (i)

Where, $v_0$is the initial velocity

To find the time t when vmax=0m/s along y axis

The final velocity along y axis is given by,

$$\begin{gathered} {\mathrm{v}}_{\mathrm{y}}={\mathrm{v}}_{0\mathrm{y}}+\mathrm{at} \\ 0=\left(12\mathrm{m}/\mathrm{s}\right)+\left(-2.0\right)\mathrm{t} \\ \mathrm{t}=6.0\mathrm{s} \end{gathered}$$

To find the velocity of the cart along x axis when it reaches itsgreatest y coordinate

The final velocity along x axis is given by

$$\begin{gathered} {\mathrm{v}}_{\mathrm{x}}={\mathrm{v}}_{0\mathrm{x}}+\mathrm{at} \\ {\mathrm{v}}_{\mathrm{x}}=\left(\frac{8.0\mathrm{m}}{\mathrm{s}}\right)+\left(\frac{4.0\mathrm{m}}{{\mathrm{s}}^2}\right)\left(6.0\mathrm{s}\right) \\ {\mathrm{v}}_{\mathrm{x}}=32\mathrm{m}/\mathrm{s} \end{gathered}$$