A particle of mass 1 kg, initially resting at origin, starts moving under the influence of a force F = 4t³ î − 3t² ĵ. If the speed of the particle at t = 1 s is √α, then the value of α is ______.

The correct answer is 2.

Step-by-step Explanation:

We are given the following values for the particle:
Mass of the particle, $m=1\text{kg}$
Force acting on the particle, $\overrightarrow{F}=4t^3\hat{i}-3t^2\hat{j}$
Initial state: resting at the origin, which means the initial velocity $\overrightarrow{v}\left(0\right)=0$ at $t=0$.

According to Newton's second law of motion, the acceleration $\overrightarrow{a}$ is related to the force and mass by:
$\overrightarrow{a}=\frac{\overrightarrow{F}}{m}$
Substituting the given force and mass into this equation:
$\overrightarrow{a}=\frac{4t^3\hat{i}-3t^2\hat{j}}{1}=4t^3\hat{i}-3t^2\hat{j}$

Since acceleration is defined as the rate of change of velocity with respect to time ($\overrightarrow{a}=\frac{d\overrightarrow{v}}{dt}$), we can write:
$d\stackrel{���}{v}=\overrightarrow{a}dt=(4t^3\hat{i}-3t^2\hat{j})dt$

Integrating both sides to find the velocity vector $\overrightarrow{v}\left(t\right)$ as a function of time:
$\overrightarrow{v}\left(t\right)=\int (4t^3\hat{i}-3t^2\hat{j})dt$
$\overrightarrow{v}\left(t\right)=t^4\hat{i}-t^3\hat{j}+\overrightarrow{C}$

Applying the initial condition that the particle is initially at rest ($\overrightarrow{v}\left(0\right)=0$ at $t=0$):
$0=0^4\hat{i}-0^3\hat{j}+\overrightarrow{C}\Rightarrow \overrightarrow{C}=0$
Thus, the velocity at any time $t$ is given by:
$\overrightarrow{v}\left(t\right)=t^4\hat{i}-t^3\hat{j}$

Now, calculate the velocity of the particle at $t=1\text{s}$:
$\overrightarrow{v}\left(1\right)=1^4\hat{i}-1^3\hat{j}=\hat{i}-\hat{j}$

The speed of the particle is the magnitude of its velocity vector:
$\text{Speed}=\left|\overrightarrow{v}\right(1\left)\right|=\sqrt{1^2+{(-1)}^2}=\sqrt{1+1}=\sqrt{2}$

We are given that the speed of the particle at $t=1\text{s}$ is $\sqrt{\alpha }$. Comparing the two values:
$\sqrt{\alpha }=\sqrt{2}$
Squaring both sides yields:
$\alpha =2$