Question

A particle starts from the origin at t=0with a velocity of $\mathbf{8}\mathbf{.}\mathbf{0}\hat{\mathbf{j}}\mathbf{m}\mathbf{/}\mathbf{s}$and moves in the x-y plane with constant acceleration $\left(4.0\hat{i}+2.0\hat{j}\right)\mathbf{m}\mathbf{/}{\mathbf{s}}^{\mathbf{2}}$. When the particle’s x-coordinate is 29 m, what are it’s (a) y-coordinate and (b) speed?

Short answer

1. The y coordinate of the particle is 45m .
2. The speed of the particle is 22m/s .

Step-by-step solution

The given data

$$\begin{gathered} \mathrm{a})\mathrm{The}\mathrm{initial}\mathrm{velocity}\mathrm{of}\mathrm{the}\mathrm{particle},({\mathrm{V}}_{\mathrm{i}})=(8\hat{\mathrm{j}})\mathrm{m}/\mathrm{s} \\ \mathrm{b})\mathrm{The}\mathrm{acceleration}\mathrm{of}\mathrm{the}\mathrm{particle},\left(\mathrm{a}\right)=(4\hat{\mathrm{i}}+2\hat{\mathrm{j}})\mathrm{m}/{\mathrm{s}}^2 \\ \mathrm{c})\mathrm{The}\mathrm{x}-\mathrm{coordinate}\mathrm{of}\mathrm{the}\mathrm{position}\mathrm{of}\mathrm{the}\mathrm{particle},(\mathrm{x})=29\mathrm{m} \end{gathered}$$

Understanding the concept of the kinematic motions

When an object is moving with a velocity, its motion is represented by a set of equations called kinematic equations. When the acceleration is constant, the kinematic equations give the relationship between the initial velocity, final velocity, acceleration, and time. Using the kinematic equations, we can find the position and speed of the particle.

$$\begin{gathered} \mathrm{Formulae}: \\ \mathrm{The}\mathrm{second}\mathrm{equation}\mathrm{of}\mathrm{kinematic}\mathrm{motion},\overrightarrow{\mathrm{r}}\left(\mathrm{t}\right)={\overrightarrow{\mathrm{V}}}_{\mathrm{i}}\mathrm{t}+\left(\frac{1}{2}\right)\overrightarrow{\mathrm{a}}{\mathrm{t}}^2\dots \left(\mathrm{i}\right) \\ \mathrm{The}\mathrm{first}\mathrm{equation}\mathrm{of}\mathrm{kinematic}\mathrm{motion},{\mathrm{V}}_{\mathrm{f}}={\mathrm{V}}_{\mathrm{i}}+\mathrm{at}\dots \left(\mathrm{ii}\right) \\ \mathrm{Magnitude}\mathrm{of}\mathrm{the}\mathrm{resultant}\mathrm{vectoris}\mathrm{given}\mathrm{by},\mathrm{V}=\sqrt{{\mathrm{V}}_{\mathrm{x}}^2+{\mathrm{V}}_{\mathrm{y}}^2}\dots \left(\mathrm{iii}\right) \end{gathered}$$

(a) Calculation of the y-coordinate of the particle

The position vector of the particle is given using equation (i) as follows:

$$\begin{gathered} \overrightarrow{\mathrm{r}}\left(\mathrm{t}\right)=\left(8\mathrm{m}/\mathrm{s}\hat{\mathrm{j}}\right)\mathrm{t}+\left(\frac{1}{2}\right)\left(4\hat{\mathrm{i}}+2\hat{\mathrm{j}}\right)\mathrm{m}/{\mathrm{s}}^2{\mathrm{t}}^2 \\ =2{\mathrm{t}}^2\hat{\mathrm{i}}+\left(8\mathrm{t}+{\mathrm{t}}^2\right)\hat{\mathrm{j}} \end{gathered}$$ …(iv)

Using the x-coordinate of the position of the particle as given, we can get the value of the time as follows:

$$\begin{gathered} 29\mathrm{m}=2{\mathrm{t}}^2 \\ \mathrm{t}=3.8\mathrm{s} \end{gathered}$$

Now, the y-coordinate of the position of the particle using the time value in equation (iv) is given as:

$$\begin{gathered} \left(\mathrm{y}\right)=8\mathrm{m}/\mathrm{s}\left(3.8\mathrm{s}\right)+\left(1.0\mathrm{m}/{\mathrm{s}}^2\right){\left(3.8\mathrm{s}\right)}^2 \\ =45\mathrm{m} \end{gathered}$$

Hence, the value of the y-coordinate is 45 m.

(b) Calculation of the speed of the particle

Using the given data in equation (ii), the final velocity of the particle is given as:

$$\begin{gathered} {\mathrm{V}}_{\mathrm{f}}=\left(8.0\mathrm{m}/\mathrm{s}\right)\hat{\mathrm{j}}+\left((4.0\mathrm{m}/{\mathrm{s}}^2\right)\hat{\mathrm{i}}+\left(2.0\mathrm{m}/{\mathrm{s}}^2\right)\hat{\mathrm{j}})\left(3.8\mathrm{s}\right) \\ =\left(15.2\mathrm{m}/\mathrm{s}\right)\hat{\mathrm{i}}+\left(15.6\mathrm{m}/\mathrm{s}\right)\overline{\mathrm{j}} \end{gathered}$$

So, the speed of the particle is given using equation (iii) as follows:

$$\begin{gathered} \mathrm{V}=\sqrt{{\left(15.2\mathrm{m}/\mathrm{s}\right)}^2+{\left(15.6\mathrm{m}/\mathrm{s}\right)}^2} \\ =22\mathrm{m}/\mathrm{s} \end{gathered}$$

Hence, the value of the speed of the particle is 22m/s