Question

An object of mass 3 kg is at rest. Now a force of \(\vec{F} = 6 t^2 \hat{i} + 4t \hat{j}\) is applied on the object then velocity of object at r = 3 second is :-

• A. $ 18 \widehat{i} + 3 \widehat{j}$
• B. $ 18 \widehat{i} + 6 \widehat{j}$
• C. $ 3 \widehat{i} + 18 \widehat{j}$
• D. $ 18 \widehat{i} + 4 \widehat{j}$

Answer 1

The Correct Option is B

To solve this problem, we need to determine the velocity of the object at \( t = 3 \) seconds when a time-dependent force is applied to it. The force vector given is: \(\vec{F} = 6 t^2 \hat{i} + 4t \hat{j}\) .

According to Newton's second law, the force acting on an object is equal to the rate of change of its linear momentum. Since the object is at rest initially, the velocity at any time can be determined by integrating the acceleration with respect to time. The acceleration vector \( \vec{a} \) is derived from the force using the relation \( \vec{F} = m \vec{a} \), where \( m \) is the mass of the object.

The mass of the object, \( m = 3 \, \text{kg} \). Therefore, the acceleration vector is:

\[ \vec{a} = \frac{\vec{F}}{m} = \frac{1}{3} (6 t^2 \hat{i} + 4t \hat{j}) = 2 t^2 \hat{i} + \frac{4t}{3} \hat{j} \]

To find the velocity \( \vec{v} \) as a function of time, we integrate the acceleration with respect to time:

\[ \vec{v}(t) = \int \vec{a} \, dt = \int (2 t^2 \hat{i} + \frac{4t}{3} \hat{j}) \, dt \] \[ \vec{v}(t) = \left( \int 2 t^2 \, dt \right) \hat{i} + \left( \int \frac{4t}{3} \, dt \right) \hat{j} \]

Calculating these integrals separately:

\[ \int 2 t^2 \, dt = \frac{2}{3} t^3 + C_1 \] \[ \int \frac{4t}{3} \, dt = \frac{2}{3} t^2 + C_2 \]

Since the object was initially at rest, the initial velocity is zero, implying \( C_1 = 0 \) and \( C_2 = 0 \).

Therefore, the velocity vector \( \vec{v}(t) \) is:

\[ \vec{v}(t) = \left( \frac{2}{3} t^3 \right) \hat{i} + \left( \frac{2}{3} t^2 \right) \hat{j} \]

Substituting \( t = 3 \) seconds:

\[ \vec{v}(3) = \left( \frac{2}{3} \times 3^3 \right) \hat{i} + \left( \frac{2}{3} \times 3^2 \right) \hat{j} \] \[ \vec{v}(3) = \left( \frac{2}{3} \times 27 \right) \hat{i} + \left( \frac{2}{3} \times 9 \right) \hat{j} \] \[ \vec{v}(3) = 18\hat{i} + 6 \hat{j} \]

Thus, the velocity of the object at \( t = 3 \) seconds is: $ 18 \hat{i} + 6 \hat{j} $.

The correct option is: $ 18 \widehat{i} + 6 \widehat{j} $.