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# Give answer! - A particle moves from the point , at t = 0, with an initial velocity . It is acted upon by a constan - Kinematics - JEE Main

A particle moves from the point $\left ( 2.0\widehat{i}+4.0\widehat{j} \right )m$, at t = 0, with an initial velocity $\left ( 5.0\widehat{i}+4.0\widehat{j} \right )ms^{-1}$. It is acted upon by a constant force which produces a constant force which produces a constant acceleration $\left ( 4.0\widehat{i}+4.0\widehat{j} \right )ms^{-2}$. What is the distance of the particle from the origin at time 2s?

• Option 1)

5 m

• Option 2)

$10\sqrt{2}m$

• Option 3)

15 m

• Option 4)

$20\sqrt{2}m$

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2nd equation or Position- time equation -

$s= ut +\frac{1}{2}at^{2}$

$s\rightarrow$ Displacement

$u\rightarrow$Initial velocity

$a\rightarrow$acceleration

$t\rightarrow$ time

-

$S=ut+\frac{1}{2}at^{2}$

t=25

$r_{i}=2\hat{i}+4\hat{j}$

$s_{i}=\vec{r}_{f}-\vec{r_{i}}$

$u=s\hat{i}+4\hat{j}$

$a=4\hat{i}+4\hat{j}$

$s=(s\hat{i}+4\hat{j})2+\frac{1}{2}\times 4\times (4\hat{i}+4\hat{j})$

$=10\hat{i}+8\hat{j}+8\hat{i}+8\hat{j}$

$s=18\hat{i}+16\hat{j}$

$\vec{r_{f}}=20\hat{i}+20\hat{j}$

$|\vec{r_{f}}|=20\sqrt{2}$

Option 1)

5 m

Option 2)

$10\sqrt{2}m$

Option 3)

15 m

Option 4)

$20\sqrt{2}m$

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