Question

Q. $v=2sin(t)\hat{i} + 2cos(t)\hat{j}$ find i) $acd^n$ at $t=\frac{\pi}{6}$ sec. ii) $\vec{s}$ at $\frac{\pi}{6}$ sec. if Particle started at t=0 sec.

Answer 1

i) $\vec{a}(\frac{\pi}{6}) = \sqrt{3}\hat{i} - \hat{j}$

ii) $\vec{s}(\frac{\pi}{6}) = (2 - \sqrt{3})\hat{i} + \hat{j}$

Answer 2

Given the velocity of the particle as a function of time: $\vec{v}(t) = 2\sin(t)\hat{i} + 2\cos(t)\hat{j}$

i) To find the acceleration $\vec{a}(t)$, we differentiate the velocity vector with respect to time: $\vec{a}(t) = \frac{d\vec{v}}{dt} = \frac{d}{dt}(2\sin(t)\hat{i} + 2\cos(t)\hat{j})$ $\vec{a}(t) = 2\cos(t)\hat{i} - 2\sin(t)\hat{j}$

Now, evaluate the acceleration at $t=\frac{\pi}{6}$ sec: $\vec{a}(\frac{\pi}{6}) = 2\cos(\frac{\pi}{6})\hat{i} - 2\sin(\frac{\pi}{6})\hat{j}$ Using $\cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2}$ and $\sin(\frac{\pi}{6}) = \frac{1}{2}$: $\vec{a}(\frac{\pi}{6}) = 2(\frac{\sqrt{3}}{2})\hat{i} - 2(\frac{1}{2})\hat{j} = \sqrt{3}\hat{i} - \hat{j}$

ii) To find the displacement $\vec{s}(t)$, we integrate the velocity vector with respect to time: $\vec{s}(t) = \int \vec{v}(t) dt = \int (2\sin(t)\hat{i} + 2\cos(t)\hat{j}) dt$ $\vec{s}(t) = (-2\cos(t) + C_1)\hat{i} + (2\sin(t) + C_2)\hat{j}$ where $C_1$ and $C_2$ are constants of integration. We can write this as $\vec{s}(t) = -2\cos(t)\hat{i} + 2\sin(t)\hat{j} + \vec{C}$, where $\vec{C} = C_1\hat{i} + C_2\hat{j}$.

The particle started at $t=0$ sec. Assuming the particle started from the origin, the displacement at $t=0$ is zero, i.e., $\vec{s}(0) = \vec{0}$. $\vec{s}(0) = (-2\cos(0) + C_1)\hat{i} + (2\sin(0) + C_2)\hat{j} = \vec{0}$ $(-2(1) + C_1)\hat{i} + (2(0) + C_2)\hat{j} = \vec{0}$ $(-2 + C_1)\hat{i} + C_2\hat{j} = 0\hat{i} + 0\hat{j}$ This implies $-2 + C_1 = 0 \implies C_1 = 2$ and $C_2 = 0$.

So, the displacement vector is: $\vec{s}(t) = (-2\cos(t) + 2)\hat{i} + (2\sin(t) + 0)\hat{j} = (2 - 2\cos(t))\hat{i} + 2\sin(t)\hat{j}$

Now, evaluate the displacement at $t=\frac{\pi}{6}$ sec: $\vec{s}(\frac{\pi}{6}) = (2 - 2\cos(\frac{\pi}{6}))\hat{i} + 2\sin(\frac{\pi}{6})\hat{j}$ Using $\cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2}$ and $\sin(\frac{\pi}{6}) = \frac{1}{2}$: $\vec{s}(\frac{\pi}{6}) = (2 - 2(\frac{\sqrt{3}}{2}))\hat{i} + 2(\frac{1}{2})\hat{j}$ $\vec{s}(\frac{\pi}{6}) = (2 - \sqrt{3})\hat{i} + \hat{j}$

Explanation:

i) Acceleration is the time derivative of velocity. Differentiate $\vec{v}(t)$ to get $\vec{a}(t)$. Substitute $t=\frac{\pi}{6}$ into $\vec{a}(t)$.

ii) Displacement is the time integral of velocity. Integrate $\vec{v}(t)$ to get $\vec{s}(t)$, including a constant of integration. Use the initial condition $\vec{s}(0)=\vec{0}$ (assuming starting from origin) to find the constant. Substitute $t=\frac{\pi}{6}$ into $\vec{s}(t)$.