Question

A particle of mass 2 kg initially at rest starts moving under the force $\vec{F} = 6t^2\hat{i} - 4at\hat{j}$.
If the speed of the particle at $t = 1$ s is $\sqrt{5}$, then the value of constant a is:

• A. 1
• B. 2
• C. 3
• D. 4

Hint:

This problem is a direct application of the integral form of Newton's second law. Remember that integration introduces a constant, which must be determined from the initial conditions of the problem. Also, distinguish carefully between velocity (a vector) and speed (the magnitude of the velocity vector).

Answer 1

The Correct Option is B

Note: Based on the context, the force vector is interpreted as $\vec{F} = 6t^2\hat{i} - 4at\hat{j}$ where 'a' is the constant to be found.
Step 1: Understanding the Question:
A particle's motion is governed by a time-dependent force. We need to find an unknown constant 'a' in the force expression, given the particle's mass, initial condition (at rest), and its speed at a specific time.
Step 2: Key Formula or Approach:
1. Use Newton's Second Law to find the acceleration: $\vec{a} = \frac{\vec{F}}{m}$.
2. Integrate the acceleration with respect to time to find the velocity: $\vec{v}(t) = \int \vec{a}(t) dt$.
3. Use the initial condition ($\vec{v}(0)=0$) to find the constant of integration.
4. Calculate the velocity vector at $t=1$ s.
5. Find the speed (magnitude of the velocity vector) and equate it to the given value to solve for 'a'.
Step 3: Detailed Explanation:
Given:
Mass, $m = 2$ kg
Force, $\vec{F}(t) = 6t^2\hat{i} - 4at\hat{j}$ N
Initial condition, $\vec{v}(0) = 0$
Speed at $t=1$ s, $|\vec{v}(1)| = \sqrt{5}$ m/s
First, find the acceleration vector:
\[ \vec{a}(t) = \frac{\vec{F}(t)}{m} = \frac{6t^2\hat{i} - 4at\hat{j}}{2} = 3t^2\hat{i} - 2at\hat{j} \text{ m/s}^2 \] Next, find the velocity vector by integrating the acceleration:
\[ \vec{v}(t) = \int \vec{a}(t) dt = \int (3t^2\hat{i} - 2at\hat{j}) dt = (\int 3t^2 dt)\hat{i} - (\int 2at dt)\hat{j} \] \[ \vec{v}(t) = t^3\hat{i} - at^2\hat{j} + \vec{C} \] Using the initial condition $\vec{v}(0) = 0$:
\[ \vec{v}(0) = (0)^3\hat{i} - a(0)^2\hat{j} + \vec{C} = 0 \implies \vec{C} = 0 \] So, the velocity vector is $\vec{v}(t) = t^3\hat{i} - at^2\hat{j}$.
Now, find the velocity at $t=1$ s:
\[ \vec{v}(1) = (1)^3\hat{i} - a(1)^2\hat{j} = 1\hat{i} - a\hat{j} \] The speed at $t=1$ s is the magnitude of this vector:
\[ |\vec{v}(1)| = \sqrt{(1)^2 + (-a)^2} = \sqrt{1 + a^2} \] We are given that this speed is $\sqrt{5}$:
\[ \sqrt{1 + a^2} = \sqrt{5} \] Squaring both sides:
\[ 1 + a^2 = 5 \] \[ a^2 = 4 \] \[ a = 2 \quad (\text{assuming a is a positive constant}) \] Step 4: Final Answer:
The value of constant a is 2.