Question

If the velocity (in \( \text{m s}^{-1} \)) of a particle at any instant \( t \) is given by \( 2.0\hat{i} + 3.0t\hat{j} \), then the magnitude of its acceleration (in \( \text{m s}^{-2} \)) is

• A. \(5 \)
• B. \(3 \)
• C. \(2 \)
• D. \(4 \)
• E. \(6 \)

Hint:

Constant components vanish on differentiation.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
Acceleration is the rate of change of velocity with respect to time. To find the acceleration vector, we need to differentiate the velocity vector with respect to time. The magnitude of the acceleration is then the length of this acceleration vector.
Step 2: Key Formula or Approach:
1. Acceleration vector: \(\vec{a} = \frac{d\vec{v}}{dt}\). 2. Magnitude of a vector \(\vec{a} = a_x\hat{i} + a_y\hat{j}\) is \(|\vec{a}| = \sqrt{a_x^2 + a_y^2}\).
Step 3: Detailed Explanation:
The velocity vector is given as: \[ \vec{v}(t) = 2.0\hat{i} + 3.0t\hat{j} \] To find the acceleration vector \(\vec{a}(t)\), we differentiate \(\vec{v}(t)\) with respect to time `t`: \[ \vec{a}(t) = \frac{d}{dt}(2.0\hat{i} + 3.0t\hat{j}) \] We differentiate component by component: \[ \vec{a}(t) = \left(\frac{d}{dt}2.0\right)\hat{i} + \left(\frac{d}{dt}3.0t\right)\hat{j} \] \[ \vec{a}(t) = 0\hat{i} + 3.0\hat{j} = 3.0\hat{j} \] The acceleration vector is constant, \(\vec{a} = 3.0\hat{j}\) ms\(^{-2}\). Now, we find the magnitude of this vector: \[ |\vec{a}| = \sqrt{0^2 + (3.0)^2} = \sqrt{9.0} = 3.0 \] The magnitude is 3.0 ms\(^{-2}\).
Step 4: Final Answer:
The magnitude of the acceleration is 3 ms\(^{-2}\).