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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Motion in a Plane
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You start from a point and your first move is 5 m at an angle of 53.1$^{\circ}$. Your second move was 6 m along the x-axis and some unknown distance along the y-axis. Your third move was some unknown distance along the x-axis and -3 m along the y-axis. If you ended up at a point 5.39 m from where you started and 21.8$^{\circ}$ below the x-axis, calculate the unknown components of your second ($y_2$) and third ($x_3$) moves .

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Answer: $x_3 = -4m,\; y_2 = -3m$
Given:
Move-1: $M_1 = 5\cos 53.1^{\circ}\hat{i} + 5 \sin 53.1^{\circ} \hat{j}= 3\hat{i}+4\hat{j}$
Move-2: $M_2 = 6\hat{i} + y_2\hat{j}$
Move-3: $M_3 = x_3\hat{i} + (-3) \hat{j} = x_3\hat{i} - 3\hat{j}$
Your final placement relative to where you started: $M_f = 5.39 \cos (-21.8^{\circ})\hat{i} + 5.39 \sin(-21.8^{\circ})\hat{j} = 5\hat{i} - 2\hat{j}$
We can now equate the individual components of the moves to their final components, i.e, $\vec{M_f} = \vec{M_1} + \vec{M_2} + \vec{M_3}$
$\Rightarrow 5 = 3 + 6 + x_3 \rightarrow x_3 = - 4$
$\Rightarrow -2 = 4 + y_2 -3 \rightarrow y_2 = -3$
answered Aug 11, 2014 by balaji.thirumalai
edited Aug 11, 2014 by balaji.thirumalai
 

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