Ncert Exercises & Solutions

4.20 The position of a particle is given by $\vec{r} = 3.0 t \hat{i} - 2.0 t^{2} \hat{j} + 4.0 \hat{k} m$

where t is in seconds and the coefficients have the proper units for $\vec{r}$ to be in meters.

(a) Find the v and a of the particle?

(b) What is the magnitude and direction of the velocity of the particle at t = 2.0 s?

NEETprep Answer:

The position of the particle is given by:

$\vec{r} = 3.0 t \hat{i} - 2.0 t^{2} \hat{j} + 4.0 \hat{k}$

Velocity

$\vec{v}$ , of the particle, is given as:

$\begin{array}{l} \vec{v} = \frac{d \vec{r}}{d t} = \frac{d}{d t} (3.0 t \hat{i} - 2.0 t^{2} \hat{j} + 4.0 \hat{k}) \\ \Rightarrow \vec{v} = 3.0 \hat{i} - 4.0 t \hat{j} \end{array}$

Acceleration

$\vec{a}$ , of the particle, is given as:

$\begin{array}{l} \vec{a} = \frac{d \vec{v}}{d t} = \frac{d}{d t} (3.0 \hat{i} - 4.0 t \hat{j}) \\ \Rightarrow \vec{a} = - 4.0 \hat{j} \end{array}$ 8.54 m/s, 69.45° below the x-axis

(b)

$\begin{array}{l} We have velocity vector, \vec{v} = 3.0 \hat{i} - 4.0 t \hat{j} \\ At t = 2.0 s \\ \vec{v} = 3.0 \hat{i} - 8.0 \hat{j} \end{array}$

The magnitude of velocity is given by:

$\begin{array}{l} | \vec{v} | = \sqrt{3^{2} + (- 8)^{2}} = \sqrt{73} = 8.54 m / s \\ \begin{matrix} Direction, θ = \tan^{- 1} (\frac{v_{y}}{v_{x}}) \\ = \tan^{- 1} (\frac{- 8}{3}) = - \tan^{- 1} (2 .667) \\ = - 69 {.45}^{\circ} \end{matrix} \end{array}$

The negative sign indicates that the direction of velocity is below the x-axis.

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