Question

A 4 kg mass moves under the influence of a force \(\vec{F} = (4t^3\hat{i} - 3t^2\hat{j})\) N where t is the time in second. If mass starts from origin at t=0, the velocity and position after t = 2 s will be:

• A. \(\vec{v} = 4\hat{i} - \frac{3}{2}\hat{j}\), \(\vec{r} = \frac{8}{5}\hat{i} - \hat{j}\)
• B. \(\vec{v} = 4\hat{i} - 2\hat{j}\), \(\vec{r} = \frac{8}{5}\hat{i} - \hat{j}\)
• C. \(\vec{v} = 4\hat{i} - 3\hat{j}\), \(\vec{r} = \frac{8}{5}\hat{i} - 2\hat{j}\)
• D. \(\vec{v} = 4\hat{i} - 3\hat{j}\), \(\vec{r} = \frac{8}{5}\hat{i} - \hat{j}\)

Hint:

For problems involving time-dependent force, the process is always F \(\rightarrow\) a \(\rightarrow\) v \(\rightarrow\) r via division by mass and then two successive integrations. Always pay close attention to the initial conditions (\(v_0\) and \(r_0\)) as they determine the constants of integration.

Answer 1

The Correct Option is B

We are given a force \(\vec{F} = (4t^3\hat{i} - 3t^2\hat{j})\) acting on a 4 kg mass. We need to determine the velocity and position of the mass at \(t = 2\) seconds, given that the mass starts from rest at the origin at\) 

First, let's apply Newton's Second Law of Motion, which states that \(\vec{F} = m\vec{a}\\) Here, the acceleration \(\vec{a}\) is:

\(\vec{a} = \frac{\vec{F}}{m} = \frac{(4t^3\hat{i} - 3t^2\hat{j})}{4} = (t^3\hat{i} - \frac{3}{4}t^2\hat{j})\)

To find the velocity, we integrate the acceleration with respect to time:

\(\vec{v}(t) = \int \vec{a} \, dt = \int (t^3\hat{i} - \frac{3}{4}t^2\hat{j}) \, dt\)

This yields:

\(\vec{v}(t) = (\frac{t^4}{4} + C_1)\hat{i} + \left(-\frac{3}{4} \cdot \frac{t^3}{3} + C_2\right)\hat{j} = \left(\frac{t^4}{4} + C_1\right)\hat{i} - \left(\frac{t^3}{4} + C_2\right)\hat{j}\)

Given that the mass starts from rest, \(\vec{v}(0) = 0\\) we have \(C_1 = 0\) and \(C_2 = 0\).

Thus, the velocity is:

\(\vec{v}(t) = \left(\frac{t^4}{4}\right)\hat{i} - \left(\frac{t^3}{4}\right)\hat{j}\)

Substitute \(t = 2\) to find the velocity at 2 seconds:

\(\vec{v}(2) = \left(\frac{2^4}{4}\right)\hat{i} - \left(\frac{2^3}{4}\right)\hat{j} = (4\hat{i} - 2\hat{j})\)

Now, to find the position, integrate the velocity with respect to time:

\(\vec{r}(t) = \int \vec{v} \, dt = \int \left(\frac{t^4}{4}\hat{i} - \frac{t^3}{4}\hat{j}\right) \, dt\)

This results in:

\(\vec{r}(t) = \left(\frac{t^5}{20} + C_3\right)\hat{i} + \left(-\frac{t^4}{16} + C_4\right)\hat{j}\)

Since the mass starts from the origin, \(\vec{r}(0) = 0\\) we have \(C_3 = 0\) and \(C_4 = 0\).

Thus, the position is:

\(\vec{r}(t) = \frac{t^5}{20}\hat{i} - \frac{t^4}{16}\hat{j}\)

Substitute \(t = 2\) to find the position at 2 seconds:

\(\vec{r}(2) = \frac{2^5}{20}\hat{i} - \frac{2^4}{16}\hat{j} = \left(\frac{32}{20}\right)\hat{i} - \left(\frac{16}{16}\right)\hat{j} = \frac{8}{5}\hat{i} - \hat{j}\)

Thus, the correct answer is:

\(\vec{v} = 4\hat{i} - 2\hat{j}\), \(\vec{r} = \frac{8}{5}\hat{i} - \hat{j}\)