Question

A particle moves in a circle of radius $R$ such that its linear speed varies with time $t$ as $v = kt$, where $k$ is a positive constant. The angle $\theta$ between the net acceleration vector and the velocity vector at time $t$ is given by:

• A. \( \tan^{-1}\left(\frac{k^2 t^2}{R}\right) \)
• B. \( \tan^{-1}\left(\frac{kt^2}{R}\right) \)
• C. \( \tan^{-1}\left(\frac{k t}{R}\right) \)
• D. \( \tan^{-1}\left(\frac{R}{k^2 t^2}\right) \)

Hint:

Test the extreme temporal boundary states to clear doubts quickly! At the absolute start (\(t = 0\)), the speed is zero, meaning there is zero centripetal turning force and the particle purely accelerates forward tangentially. Thus, the angle \(\theta\) must be \(0^\circ\). Substituting \(t = 0\) into our options shows that \(\tan^{-1}(0) = 0^\circ\) for Option B, verifying the physical validity of our formula.

Answer 1

The Correct Option is B

Concept: The velocity vector \(\vec{v}\) of a particle in circular motion is always directed tangentially along the path. Therefore, the angle \(\theta\) between the net acceleration vector \(\vec{a}_{\text{total}}\) and the velocity vector is exactly the angle between \(\vec{a}_{\text{total}}\) and the tangential acceleration component \(\vec{a}_t\). Using vector geometry, the tangent of this angle is the ratio of the perpendicular acceleration component (centripetal acceleration, \(a_c\)) to the base component (tangential acceleration, \(a_t\)): \[ \tan\theta = \frac{a_c}{a_t} \quad \implies \quad \theta = \tan^{-1}\left(\frac{a_c}{a_t}\right) \]

Step 1: Calculating the tangential acceleration component ($a_t$).
Tangential acceleration is defined as the rate of change of linear speed: \[ a_t = \frac{dv}{dt} = \frac{d}{dt}(kt) = k \]

Step 2: Calculating the centripetal acceleration component ($a_c$).
Centripetal acceleration is determined by the instantaneous linear speed and path radius: \[ a_c = \frac{v^2}{R} = \frac{(kt)^2}{R} = \frac{k^2 t^2}{R} \]

Step 3: Evaluating the direction angle $\theta$.
Substitute both derived component expressions into our trigonometric ratio layout: \[ \tan\theta = \frac{\left(\frac{k^2 t^2}{R}\right)}{k} = \frac{k^2 t^2}{R \cdot k} = \frac{kt^2}{R} \] Isolating the angle yields the final expression: \[ \theta = \tan^{-1}\left(\frac{kt^2}{R}\right) \]