An object of mass 3 kg is at rest. Now a force of $\overrightarrow{\mathrm{F}}=6{\mathrm{t}}^2\widehat{\mathrm{i}}+4\mathrm{t}\widehat{\mathrm{j}}$ is applied on the object then velocity of object at t = 3 sec is

Given that an object of mass 3 kg is at rest, and a force of F = 6t² î + 4t ĵ is applied on the object, we are tasked to find the velocity of the object at t = 3 seconds.

Step 1: Identify the given parameters

• Mass of the object: m = 3 kg
• Applied force: F = 6t² î + 4t ĵ

Step 2: Calculate the acceleration

We will use Newton's second law, which states that the force applied on an object is equal to the product of its mass and acceleration, i.e.,

F = m * a

Substitute the given values:

6t² î + 4t ĵ = 3 * a

Solving for acceleration a, we get:

a = (6t² î + 4t ĵ) / 3

Which simplifies to:

a = 2t² î + (4/3)t ĵ

Step 3: Relate acceleration to velocity

Acceleration is the derivative of velocity with respect to time:

a = dv/dt

This means:

dv/dt = 2t² î + (4/3)t ĵ

Step 4: Integrate to find velocity

To find velocity, we integrate both sides with respect to time t:

v = ∫(2t² î + (4/3)t ĵ) dt

Performing the integration on each term:

v = (2/3)t³ î + (2/3)t² ĵ + C

Here, C is the constant of integration. Since the object is at rest initially, the initial velocity is zero, which gives C = 0. Therefore, the velocity equation becomes:

v = (2/3)t³ î + (2/3)t² ĵ

Step 5: Calculate velocity at t = 3 seconds

Substitute t = 3 into the velocity equation:

v = (2/3)(3³) î + (2/3)(3²) ĵ

Now, calculate each term:

v = (2/3)(27) î + (2/3)(9) ĵ

v = 18 î + 6 ĵ

Final Answer:

The velocity of the object at t = 3 seconds is:

v = 18 î + 6 ĵ m/s