Question

Q. $v = 2sin(t)\hat{i} + 2cos(t)\hat{j}$ find i) $accl^n$ at $t = \frac{\pi}{6}$ sec.

ii) $\overrightarrow{S}$ at $\frac{\pi}{6}$ sec sec.

if Particle started at t=0sec.

Answer 1

i) $\sqrt{3}\hat{i} - \hat{j}$ ii) $(2 - \sqrt{3})\hat{i} + \hat{j}$

Answer 2

i) Acceleration is the time derivative of velocity. Differentiate the given velocity vector $\vec{v}(t)$ with respect to $t$ to get $\vec{a}(t)$. Substitute $t = \frac{\pi}{6}$ into $\vec{a}(t)$ and evaluate the trigonometric functions $\cos(\frac{\pi}{6})$ and $\sin(\frac{\pi}{6})$ to find the acceleration vector at that time.

ii) Displacement is the change in position. The displacement vector from time $t_1$ to $t_2$ is the definite integral of the velocity vector from $t_1$ to $t_2$. Here, $t_1 = 0$ and $t_2 = \frac{\pi}{6}$. Integrate the given velocity vector $\vec{v}(t)$ with respect to $t$ from $0$ to $\frac{\pi}{6}$. Evaluate the definite integral by substituting the limits and subtracting the value at the lower limit from the value at the upper limit. This gives the displacement vector at $t=\frac{\pi}{6}$ sec relative to the position at $t=0$ sec.