Question

A particle moves along a semi circular path of radius $r$ in time $t$ with constant speed, then its

• A. average speed is $\frac{\pi r}{2t}$
• B. average velocity is $\frac{r}{t}$
• C. displacement is $2\pi r$
• D. distance travelled is $2r$
• E. average acceleration is $\frac{2\pi r}{t^{2}}$

Hint:

Logic Tip: A common mistake is assuming average acceleration is zero because "speed is constant." Acceleration measures the change in the velocity *vector*. A complete $180^{\circ}$ turn creates the maximum possible velocity change ($2v$) for a given speed.

Answer 1

Answer: (E)

Concept:
Kinematics of circular motion involves distinguishing between scalar and vector quantities. Distance is the actual path length, while displacement is the shortest straight-line distance between initial and final points. Average velocity = $\frac{\text{Displacement}}{\text{Time}}$, and Average acceleration = $\frac{\text{Change in Velocity}}{\text{Time}}$.
Step 1: Analyze the distance and displacement.
The particle travels along a semi-circle (half of a full circle). Total distance traveled $d = \frac{1}{2}(2\pi r) = \pi r$. (Option D is incorrect). The displacement vector connects the start and end points directly across the diameter. Magnitude of displacement $|\vec{s}| = 2r$. (Option C is incorrect).
Step 2: Analyze the average speed and velocity.
Average speed $= \frac{\text{Distance}}{t} = \frac{\pi r}{t}$. (Option A is incorrect). Average velocity magnitude $= \frac{\text{Displacement}}{t} = \frac{2r}{t}$. (Option B is incorrect).
Step 3: Analyze the average acceleration.
Average acceleration is a vector quantity: $\vec{a}_{avg} = \frac{\vec{v}_{final} - \vec{v}_{initial}}{t}$. The particle moves with constant speed $v$. However, its direction changes continuously. In a semi-circle, the final velocity vector is exactly anti-parallel (opposite direction) to the initial velocity vector. If $\vec{v}_{initial} = v\hat{i}$, then $\vec{v}_{final} = -v\hat{i}$. Change in velocity $\Delta \vec{v} = (-v\hat{i}) - (v\hat{i}) = -2v\hat{i}$. Magnitude of change in velocity $|\Delta \vec{v}| = 2v$. Substitute the speed $v = \frac{\pi r}{t}$ into this expression: $|\Delta \vec{v}| = 2\left(\frac{\pi r}{t}\right)$ Now calculate the magnitude of the average acceleration: $$|\vec{a}_{avg}| = \frac{|\Delta \vec{v}|}{t} = \frac{\frac{2\pi r}{t}}{t} = \frac{2\pi r}{t^2}$$ This matches Option E.