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Question: A point traversed 3/4th of the circle of radius R in time t. The magnitude of the average velocity o...

A point traversed 3/4th of the circle of radius R in time t. The magnitude of the average velocity of the particle in this time interval is :

A

πRt\frac{\pi R}{t}

B

3πR2t\frac{3\pi R}{2t}

C

R2t\frac{R\sqrt{2}}{t}

D

R2t\frac{R}{\sqrt{2}t}

Answer

R2t\frac{R\sqrt{2}}{t}

Explanation

Solution

The problem asks for the magnitude of the average velocity of a particle that traverses 3/4th of a circle of radius R in time t.

  1. Understanding Average Velocity:
    Average velocity is defined as the total displacement divided by the total time taken.
    v⃗avg=Δr⃗Δt\vec{v}_{avg} = \frac{\Delta \vec{r}}{\Delta t}
    The magnitude of the average velocity is ∣v⃗avg∣=∣Δr⃗∣Δt|\vec{v}_{avg}| = \frac{|\Delta \vec{r}|}{\Delta t}.

  2. Determining Displacement (Δr⃗\Delta \vec{r}):
    Let the circle be centered at the origin (0,0).
    Assume the particle starts at point A on the positive x-axis, so its initial position vector is r⃗A=Ri^\vec{r}_A = R\hat{i}.
    The particle traverses 3/4th of the circle. This corresponds to an angular displacement of 34×2π=3π2\frac{3}{4} \times 2\pi = \frac{3\pi}{2} radians (or 270 degrees).
    If the particle moves counter-clockwise from (R,0)(R, 0):

  • After 90∘90^\circ (1/4 circle), it would be at (0,R)(0, R).
  • After 180∘180^\circ (1/2 circle), it would be at (−R,0)(-R, 0).
  • After 270∘270^\circ (3/4 circle), it would be at (0,−R)(0, -R).

So, the final position vector is r⃗B=−Rj^\vec{r}_B = -R\hat{j}.

The displacement vector Δr⃗\Delta \vec{r} is the vector from the initial position to the final position:
Δr⃗=r⃗B−r⃗A=(−Rj^)−(Ri^)=−Ri^−Rj^\Delta \vec{r} = \vec{r}_B - \vec{r}_A = (-R\hat{j}) - (R\hat{i}) = -R\hat{i} - R\hat{j}

The magnitude of the displacement ∣Δr⃗∣|\Delta \vec{r}| is the length of this vector:
∣Δr⃗∣=(−R)2+(−R)2=R2+R2=2R2=R2|\Delta \vec{r}| = \sqrt{(-R)^2 + (-R)^2} = \sqrt{R^2 + R^2} = \sqrt{2R^2} = R\sqrt{2}

Alternatively, visualize the initial point (R,0)(R,0) and the final point (0,−R)(0,-R). These two points form a right-angled isosceles triangle with the center of the circle (origin). The two equal sides are the radii R, and the displacement is the hypotenuse. By Pythagorean theorem, the length of the hypotenuse is R2+R2=2R2=R2\sqrt{R^2 + R^2} = \sqrt{2R^2} = R\sqrt{2}.

  1. Determining Time Taken (Δt\Delta t):
    The time taken is given as tt.

  2. Calculating the Magnitude of Average Velocity:
    Now, substitute the values into the formula for the magnitude of average velocity:
    ∣v⃗avg∣=∣Δr⃗∣Δt=R2t|\vec{v}_{avg}| = \frac{|\Delta \vec{r}|}{\Delta t} = \frac{R\sqrt{2}}{t}