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# An object starts from rest, and moves under the acceleration $a=4i$ its position after 3s is given by $r=7i+4j$ what is its initial position?

Last updated date: 09th Aug 2024
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Hint: First of all we will use a displacement formula in which we will substitute value of displacement $\left( \overrightarrow{S} \right)$ with $\overrightarrow{S}=$ final position – initial position. After doing that we can find the initial position of an object.
Formula used: $\overrightarrow{S}=\overrightarrow{u}t+\dfrac{1}{2}\overrightarrow{a}{{t}^{2}}$

Given data
\begin{align} & \overrightarrow{a}=4\widehat{i} \\ & t=3s \\ & \overrightarrow{r}=7\widehat{i}+4\widehat{j} \\ \end{align}
Here $\overrightarrow{r}$ is the final position so let’s assume initial position is$\overrightarrow{q}$,
Now displacement $\left( \overrightarrow{S} \right)$
$\overrightarrow{S}=$ Final position – initial position
$\overrightarrow{S}=\overrightarrow{r}-\overrightarrow{q}.....(1)$
Now formula for displacement as equation of motion,
$\overrightarrow{S}=\overrightarrow{u}t+\dfrac{1}{2}\overrightarrow{a}{{t}^{2}}.....\left( 2 \right)$
Where, $\overrightarrow{S}=$ displacement of the particle
$\overrightarrow{u}$= initial velocity of the particle
t = time
$\overrightarrow{a}$ = acceleration of the particle
Here the object start moving from the rest hence its initial velocity is zero $\overrightarrow{u}=0$
Now substitute the value of equation (1) into equation (2)
\begin{align} & \Rightarrow \overrightarrow{r}-\overrightarrow{q}=\left( 0 \right)\times \left( 3 \right)+\dfrac{1}{2}(4\widehat{i}){{(3)}^{2}} \\ & \Rightarrow 7\widehat{i}+4\widehat{j}-\overrightarrow{q}=\dfrac{1}{2}\times 36\widehat{i} \\ & \Rightarrow -\overrightarrow{q}=18\widehat{i}-7\widehat{i}-4\widehat{j} \\ & \Rightarrow -\overrightarrow{q}=+11\widehat{i}-4\widehat{j} \\ & \therefore \overrightarrow{q}=-11\widehat{i}+4\widehat{j} \\ \end{align}
Therefore initial position of particle is $\overrightarrow{q}=-11\widehat{i}+4\widehat{j}$