
A particle moves from the point $\left( {2.0\hat i + 4.0\hat j} \right)m$ at $t = 0$, with an initial velocity $\left( {5.0\hat i + 4.0\hat j} \right)m{s^{ - 1}}$. It is acted upon by a constant force which produces a constant acceleration $\left( {4.0\hat i + 4.0\hat j} \right)m{s^{ - 2}}$. What is the distance of the particle from the origin at time $2s$ ?
A. $20\sqrt 2 m$
B. $10\sqrt 2 m$
C. $5m$
D. $15m$
Answer
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Hint: To solve the question we will use the second equation of motion in vector form $\vec s = \vec ut + \dfrac{1}{2}\vec a{t^2}$ to and then equate with difference of final and initial position will give the distance and then find magnitude by $|\vec A| = \sqrt {A_x^2 + A_y^2 + A_z^2} $.
Complete step by step answer:
Now from the question
Given initial position vector \[{\vec r_i} = \left( {2.0\hat i + 4.0\hat j} \right)m\]
Initial velocity vector \[\vec u = \left( {5.0\hat i + 4.0} \right)m{s^{ - 1}}\]
Acceleration vector $\vec a = \left( {4.0\hat i + 4.0\hat j} \right)m{s^{ - 2}}$
Now from second equation of motion we have
$\vec s = \vec ut + \dfrac{1}{2}\vec a{t^2}$
Now $\vec s = \left( {5.0\hat i + 4.0\hat j} \right)\left( 2 \right) + \dfrac{1}{2}\left( {4.0\hat i + 4.0\hat j} \right){\left( 2 \right)^2}$
\[\vec s = \left( {18\hat i + 16\hat j} \right)\]
Now the distance covered by particle from origin at $2s$
\[\begin{gathered}
{{\vec r}_f} - {{\vec r}_i} = \vec s \\
{{\vec r}_f} - \left( {2.0\hat i + 4.0\hat j} \right) = \left( {18\hat i + 16\hat j} \right) \\
{{\vec r}_f} = \left( {18\hat i + 16\hat j} \right) + \left( {2.0\hat i + 4.0\hat j} \right) \\
\end{gathered} \]
${\vec r_f} = \left( {20\hat i + 20\hat j} \right)$
Magnitude of distance:
$|\vec A| = \sqrt {A_x^2 + A_y^2 + A_z^2} $
Hence
$
|{{\vec r}_f}| = \sqrt {{{\left( {20} \right)}^2} + {{\left( {20} \right)}^2}} \\
|{{\vec r}_f}| = \sqrt {400 + 400} = \sqrt {800} \\
|{{\vec r}_f}| = 20\sqrt 2 m \\
$
Therefore the correct option is (A).
Note:
In such types of questions first note down whatever is given like in the above question we have a point from where a particle moves and its having an initial velocity and after that a constant force acted upon that so to obtain the distance of the particle from the origin we first find out the vector and after that the magnitude of the vector.
Complete step by step answer:
Now from the question
Given initial position vector \[{\vec r_i} = \left( {2.0\hat i + 4.0\hat j} \right)m\]
Initial velocity vector \[\vec u = \left( {5.0\hat i + 4.0} \right)m{s^{ - 1}}\]
Acceleration vector $\vec a = \left( {4.0\hat i + 4.0\hat j} \right)m{s^{ - 2}}$
Now from second equation of motion we have
$\vec s = \vec ut + \dfrac{1}{2}\vec a{t^2}$
Now $\vec s = \left( {5.0\hat i + 4.0\hat j} \right)\left( 2 \right) + \dfrac{1}{2}\left( {4.0\hat i + 4.0\hat j} \right){\left( 2 \right)^2}$
\[\vec s = \left( {18\hat i + 16\hat j} \right)\]
Now the distance covered by particle from origin at $2s$
\[\begin{gathered}
{{\vec r}_f} - {{\vec r}_i} = \vec s \\
{{\vec r}_f} - \left( {2.0\hat i + 4.0\hat j} \right) = \left( {18\hat i + 16\hat j} \right) \\
{{\vec r}_f} = \left( {18\hat i + 16\hat j} \right) + \left( {2.0\hat i + 4.0\hat j} \right) \\
\end{gathered} \]
${\vec r_f} = \left( {20\hat i + 20\hat j} \right)$
Magnitude of distance:
$|\vec A| = \sqrt {A_x^2 + A_y^2 + A_z^2} $
Hence
$
|{{\vec r}_f}| = \sqrt {{{\left( {20} \right)}^2} + {{\left( {20} \right)}^2}} \\
|{{\vec r}_f}| = \sqrt {400 + 400} = \sqrt {800} \\
|{{\vec r}_f}| = 20\sqrt 2 m \\
$
Therefore the correct option is (A).
Note:
In such types of questions first note down whatever is given like in the above question we have a point from where a particle moves and its having an initial velocity and after that a constant force acted upon that so to obtain the distance of the particle from the origin we first find out the vector and after that the magnitude of the vector.
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