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Step-by-Step Solution
Step 1: Determine the Angular Velocity
The body completes one revolution (angle $2\pi$) in $4\,\text{s}$. The angular velocity $ \omega $ is given by:
$ \omega = \frac{2\pi}{T} = \frac{2\pi}{4} = \frac{\pi}{2}\,\text{rad/s}.$
Step 2: Calculate the Angular Displacement at 3 seconds
In uniform circular motion, the angular displacement $ \theta $ after time $t$ is:
$ \theta = \omega \, t.$
For $ t = 3 \,\text{s} $:
$ \theta = \frac{\pi}{2} \times 3 = \frac{3\pi}{2} \,\text{radians}.$
Step 3: Determine the Coordinates of the Body at 3 seconds
Let the radius of the circle be $ r = 10 \,\text{m} $. If the body started on the positive $x$-axis, then its coordinates at an angle $ \theta $ from the starting point can be written as:
$ (x,\,y) = \bigl(r \cos(\theta),\, r \sin(\theta)\bigr). $
Substituting $ \theta = \tfrac{3\pi}{2} $:
$ x = 10 \cos\!\bigl(\tfrac{3\pi}{2}\bigr) = 10 \times 0 = 0, $
$ y = 10 \sin\!\bigl(\tfrac{3\pi}{2}\bigr) = 10 \times (-1) = -10. $
Thus, at $ t = 3\,\text{s}, $ the body is at $ (0,\,-10). $
Step 4: Calculate the Displacement from the Starting Point
The body starts at $ (10,\,0). $ So, the displacement vector from $ (10,\,0) $ to $ (0,\,-10) $ is:
$ \langle 0 - 10,\, -10 - 0 \rangle = \langle -10,\,-10 \rangle. $
The magnitude of this displacement is:
$ d = \sqrt{(-10)^2 + (-10)^2} = \sqrt{100 + 100} = \sqrt{200} = 10\sqrt{2}. $
Step 5: Final Answer
Therefore, at the end of 3 seconds, the displacement of the body from its starting point is
$ 10\sqrt{2}\,\text{m}. $
