Question

A particle leaves the origin with an initial velocity$\overrightarrow{\mathbf{v}}\mathbf{=}\left(3.00\right)\mathbf{}\hat{\mathbf{i}}\mathbf{}\mathbf{m}\mathbf{/}\mathbf{s}$and a constant acceleration$\overrightarrow{\mathbf{a}}\mathbf{=}\left(‐1.00\hat{i}‐0.500\hat{j}\right)\mathbf{}\mathbf{m}\mathbf{/}{\mathbf{s}}^{\mathbf{2}}$. When it reaches its maximum xcoordinate, what are its: (A) velocity and (b) position vector.

Short answer

(a) The velocity is$\mathrm{v}=‐1.5\mathrm{m}/\mathrm{s}\hat{\mathrm{j}}$

(b) Position vector is$\overline{\mathrm{r}}=\left(4.45\mathrm{m}\right)\hat{\mathrm{i}}-\left(2.25\mathrm{m}\right)\hat{\mathrm{j}}$

Step-by-step solution

Given information

It is given that,

$$\begin{gathered} \overrightarrow{\mathrm{v}}=3.00\hat{\mathrm{i}} \\ \overrightarrow{\mathrm{a}}=-1.00\hat{\mathrm{i}}-0.5\hat{\mathrm{j}} \end{gathered}$$

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 can be used to find the velocity and the position vector of a particle.To find velocity, acceleration can be found in x and y direction fromthegiven vector. And further the position vector can be computed.

Formulae:

The final velocity in kinematic equation can be written as,

$v_f=v_0+at$

Where $v_0$is the initial velocity.

The displacement is given by,

$x=v_0\times t+\frac{1}{2}a{t}^2$

(a) To find the velocity vector

Using equation (i) the velocity vector along x direction is given by,

$v_{fx}=v_{0x}+a_{x}t$

When particle reaches maximum x coordinate then$v_{xf}=0$

So,$0=3-1\times t$

$t=3s$

Nowthecomponent of velocity in y direction is given by,

$$\begin{gathered} v_{fy}=v_{0y}+a_{y}t \\ {v}_{fy}=0-0.5\times 3 \\ {v}_{fy}=-1.5\mathrm{m}/\mathrm{s} \end{gathered}$$

Thus, velocity vector at maximum x is as follow:

$\overrightarrow{v}=\left(-1.5\mathrm{m}/\mathrm{s}\right)\hat{\mathrm{j}}$

(b) To find the position vector

Components of position vectors can be given by the following equation:

$$\begin{gathered} x=v_{0x}t+\frac{1}{2}{a}_x{t}^2 \\ x=3\times 3+\frac{1}{2}\left(-1\right)\times 3^2 \\ x=4.5\mathrm{m} \end{gathered}$$

Now, y component of position vector is as follow:

$$\begin{gathered} y=v_{0y}t+\frac{1}{2}{a}_y{t}^2 \\ y=0+\frac{1}{2}\left(-0.5\right)3^2 \\ y=-2.25\mathrm{m} \end{gathered}$$

So unit vector is given as,

$\overrightarrow{r}=\left(4.5\mathrm{m}\right)\hat{\mathrm{i}}-\left(2.25\mathrm{m}\right)\hat{\mathrm{j}}$