First slide

Rectilinear Motion

Question

A particle moves from the point $(2 \hat{i} + 4 \hat{j}) m$ , at t=0 with an initial velocity $(5 \hat{i} + 4 \hat{j}) m / s$ . It is acted upon by a constant acceleration $(4 \hat{i} + 4 \hat{j}) m / s^{2}$ . What is the distance of the particle from the origin at time $t = 2 s$ in meters?

Moderate

Solution

Given
Initial Position, ${\vec{r}}_{o} = (2 \hat{i} + 4 \hat{j}) m$
Initial velocity, $\vec{u} = (5 \hat{i} + 4 \hat{j}) m / s$
Acceleration, $\vec{a} = (4 \hat{i} + 4 \hat{j}) m / s^{2}$
$\begin{array}{l} \vec{S} = \vec{u} t + \frac{1}{2} a \vec{} t^{2} \\ \Rightarrow \vec{r} - {\vec{r}}_{o} = (5 \hat{i} + 4 \hat{j}) \times 2 + \frac{1}{2} (4 \hat{i} + 4 \hat{j}) \times 4 \\ \Rightarrow \vec{r} = (2 \hat{i} + 4 \hat{j}) + (10 \hat{i} + 8 \hat{j}) + 8 \hat{i} + 8 \hat{j} \\ \Rightarrow \vec{r} = 20 \hat{i} + 20 \hat{j} \\ \Rightarrow | \vec{r} | = 20 \sqrt{2} m \\ \Rightarrow | \vec{r} | = 28.28 m \end{array}$

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