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# The mosquito net over a $7\;{\text{ft}} \times 4\;{\text{ft}}$bed is $3\;{\text{ft}}$ high. The net has a hole at one corner of the bed through which a mosquito enters the net. Find the magnitude of the displacement of the mosquito.A) $\sqrt {54} \;{\text{ft}}$ B) $81\;{\text{ft}}$ C) $\sqrt {74} \;{\text{ft}}$ D) $\sqrt {64} \;{\text{ft}}$

Last updated date: 19th Sep 2024
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Hint: The above problem is based on the kinematics. The displacement is defined as the change in the position and direction of the object. The net over the bed is the same as the box in the shape of a cube. The displacement of the mosquito can be found by the formula to find the length of the diagonal of the box.

Given: The length of the bed is $l = 7\;{\text{ft}}$, width of the bed is $b = 4\;{\text{ft}}$ and height of the net is $h = 3\;{\text{ft}}$.
$d = \sqrt {{l^2} + {b^2} + {h^2}}$
Substitute $7\;{\text{ft}}$for l, $4\;{\text{ft}}$for b and $3\;{\text{ft}}$for h in the above formula to find the magnitude of the displacement of the mosquito.
$d = \sqrt {{{\left( {7\;{\text{ft}}} \right)}^2} + {{\left( {4\;{\text{ft}}} \right)}^2} + {{\left( {3\;{\text{ft}}} \right)}^2}}$
$d = \sqrt {74} \;{\text{ft}}$
Thus, the magnitude of the displacement of the mosquito is $\sqrt {74} \;{\text{ft}}$and the