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Question: A particle moves from the point \(\left( {2.0\hat i + 4.0\hat j} \right)m\) at \(t = 0\), with an in...

A particle moves from the point (2.0i^+4.0j^)m\left( {2.0\hat i + 4.0\hat j} \right)m at t=0t = 0, with an initial velocity (5.0i^+4.0j^)ms1\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 (4.0i^+4.0j^)ms2\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 2s2s ?
A. 202m20\sqrt 2 m
B. 102m10\sqrt 2 m
C. 5m5m
D. 15m15m

Explanation

Solution

To solve the question we will use the second equation of motion in vector form s=ut+12at2\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 A=Ax2+Ay2+Az2|\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 ri=(2.0i^+4.0j^)m{\vec r_i} = \left( {2.0\hat i + 4.0\hat j} \right)m
Initial velocity vector u=(5.0i^+4.0)ms1\vec u = \left( {5.0\hat i + 4.0} \right)m{s^{ - 1}}
Acceleration vector a=(4.0i^+4.0j^)ms2\vec a = \left( {4.0\hat i + 4.0\hat j} \right)m{s^{ - 2}}
Now from second equation of motion we have
s=ut+12at2\vec s = \vec ut + \dfrac{1}{2}\vec a{t^2}
Now s=(5.0i^+4.0j^)(2)+12(4.0i^+4.0j^)(2)2\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}
s=(18i^+16j^)\vec s = \left( {18\hat i + 16\hat j} \right)
Now the distance covered by particle from origin at 2s2s

{{\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.