Question

A particle moves so that its position (in meters) as a function of time (in seconds) is $\stackrel{\mathbf{\rightharpoonup }}{\mathbf{r}}\mathbf{=}\hat{\mathbf{i}}\mathbf{+}\mathbf{4}{\mathbf{t}}^{\mathbf{2}}\hat{\mathbf{j}}\mathbf{+}\mathbf{t}\hat{\mathbf{k}}$. Write expressions for: (a) its velocity and (b) its acceleration as functions of time.

Short answer

(a) The velocity is$(\left(8\mathrm{t}\right)\hat{\mathrm{j}}+\widehat{\mathrm{k})}\mathrm{m}/\mathrm{s}$

(b) The acceleration is $8\hat{\mathrm{j}}\mathrm{m}/{\mathrm{s}}^2$

Step-by-step solution

Given information

It is given that,

$\stackrel{\rightharpoonup }{\mathrm{r}}=\hat{\mathrm{i}}+4{\mathrm{t}}^2\hat{\mathrm{j}}+\mathrm{t}\hat{\mathrm{k}}$

To understand the concept

This problem deals with simple algebraic operation that involves calculation of velocity and the acceleration at given time. Velocity is the rate of change of its position and the acceleration is the rate of change of velocity with respect to time. Thus the velocity vector and acceleration can be expressed using the standard equation for the velocity and acceleration.

Formulae:

$$\begin{gathered} \stackrel{\rightharpoonup }{\mathrm{v}}=\frac{\mathrm{d}\stackrel{\rightharpoonup }{\mathrm{r}}}{\mathrm{dt}}\left(i\right) \\ \mathrm{Where},\stackrel{\rightharpoonup }{\mathrm{r}}\mathrm{is}\mathrm{the}\mathrm{position}\mathrm{vector} \\ \stackrel{\rightharpoonup }{\mathrm{a}}=\frac{\mathrm{d}\stackrel{\rightharpoonup }{\mathrm{v}}}{\mathrm{dt}}\left(ii\right) \end{gathered}$$

(a) To find the velocity

Velocity can be expressed by substituting the value of $\stackrel{\rightharpoonup }{r}$in equation (i)

$\stackrel{\rightharpoonup }{\mathrm{v}}=\frac{\mathrm{d}\left(\hat{\mathrm{i}}+4{\mathrm{t}}^2\hat{\mathrm{j}}+\mathrm{t}\hat{\mathrm{k}}\right)}{\mathrm{dt}}$

Thus,

$\stackrel{\rightharpoonup }{\mathrm{v}}=\left(\left(8\mathrm{t}\right)\hat{\mathrm{j}}+\hat{\mathrm{k}}\right)\mathrm{m}/\mathrm{s}$

(b) To find the acceleration

Acceleration can be expressed by substituting the value of $\stackrel{\rightharpoonup }{v}$in equation (ii)

$\stackrel{\rightharpoonup }{\mathrm{a}}=\frac{\mathrm{d}(\left(8\mathrm{t}\right)\hat{\mathrm{i}}+\hat{\mathrm{k}})}{\mathrm{dt}}$

Thus, the acceleration is

$\stackrel{\rightharpoonup }{\mathrm{a}}=8\hat{\mathrm{j}}\mathrm{m}/{\mathrm{s}}^2$