
A particle has a displacement of 12 m towards east then 5 m towards north and then 6 m vertically upward. The resultant displacement is:
A) 10.04 m
B) 12.10 m
C) 14.32 m
D) 12.6 m
Answer
233.1k+ views
Hint: In this question we have to find the resultant displacement of a particle moving in vertically upward direction. Initially the particle has a displacement of 12 m towards east and then it moves 5 m towards north. It shows that the initial displacements are perpendicular to each other.
Complete step by step solution:
Given,
${d_1} = 12m$
${d_2} = 5m$
${d_3} = 6m$
Now, we will find the resultant displacement of the particle due to its motion of 12 m towards east and 5 m towards north. Let the resultant displacement is ${R_1}$ then,
$\Rightarrow {R_1} = \sqrt {{d_1}^2 + {d_2}^2} $
Putting the values of ${d_1}$and ${d_2}$
$\Rightarrow {R_1} = \sqrt {{{(12)}^2} + {{(5)}^2}} $
$\Rightarrow {R_1} = \sqrt {144 + 25} $
$\Rightarrow {R_1} = \sqrt {169} $
$\Rightarrow {R_1} = 13m$
Now, the motion is 6 m in vertically upward direction. Therefore, the angle between ${R_1}$ and ${d_3}$ will be perpendicular to each other. Let the resultant displacement is ${R_2}$ then,
$\Rightarrow {R_2} = \sqrt {{R_1}^2 + {d_3}^2} $
Putting the values of ${R_1}$ and ${d_3}$
$\Rightarrow {R_2} = \sqrt {{{(13)}^2} + {{(6)}^2}} $
$\Rightarrow {R_2} = \sqrt {169 + 36} $
\[\Rightarrow {R_2} = \sqrt {205} \]
\[\Rightarrow {R_2} = 14.32m\]
Hence, from above calculation we have seen that the resultant displacement of the motion is \[{R_2} = 14.32m\].
Note: In the given question we have used the concept of distance between two points, when the points are at right angles to each other. In the given question or in such types of questions we have to be careful about the direction we are finding. Because here it was quite easy to follow that the angle between east and north direction will be $90^\circ $. After this the resultant of this motion will also be perpendicular to the motion in vertically upward direction. So, we have to be careful about the direction and angle. One more thing that we should be careful about is that the unit of all the displacements is in the same system. So, before starting calculation we should check the unit of all the distances.
Complete step by step solution:
Given,
${d_1} = 12m$
${d_2} = 5m$
${d_3} = 6m$
Now, we will find the resultant displacement of the particle due to its motion of 12 m towards east and 5 m towards north. Let the resultant displacement is ${R_1}$ then,
$\Rightarrow {R_1} = \sqrt {{d_1}^2 + {d_2}^2} $
Putting the values of ${d_1}$and ${d_2}$
$\Rightarrow {R_1} = \sqrt {{{(12)}^2} + {{(5)}^2}} $
$\Rightarrow {R_1} = \sqrt {144 + 25} $
$\Rightarrow {R_1} = \sqrt {169} $
$\Rightarrow {R_1} = 13m$
Now, the motion is 6 m in vertically upward direction. Therefore, the angle between ${R_1}$ and ${d_3}$ will be perpendicular to each other. Let the resultant displacement is ${R_2}$ then,
$\Rightarrow {R_2} = \sqrt {{R_1}^2 + {d_3}^2} $
Putting the values of ${R_1}$ and ${d_3}$
$\Rightarrow {R_2} = \sqrt {{{(13)}^2} + {{(6)}^2}} $
$\Rightarrow {R_2} = \sqrt {169 + 36} $
\[\Rightarrow {R_2} = \sqrt {205} \]
\[\Rightarrow {R_2} = 14.32m\]
Hence, from above calculation we have seen that the resultant displacement of the motion is \[{R_2} = 14.32m\].
Note: In the given question we have used the concept of distance between two points, when the points are at right angles to each other. In the given question or in such types of questions we have to be careful about the direction we are finding. Because here it was quite easy to follow that the angle between east and north direction will be $90^\circ $. After this the resultant of this motion will also be perpendicular to the motion in vertically upward direction. So, we have to be careful about the direction and angle. One more thing that we should be careful about is that the unit of all the displacements is in the same system. So, before starting calculation we should check the unit of all the distances.
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