Question

What is the instantaneous velocity of an object moving in accordance to `f(t)= (sin(t+pi),sin(2t-pi/4))` at `t=(-pi)/3`?

Answer 1

Given equation can be re-written as a x as a function of time in a cartesian system, that is
`x(t)=sin(t+\pi)\hat{i}+sin(2t-\pi/4)\hat{j}`
Differentiating with respect to time, we get the instantaneous velocity of the object at time `t`,
So, `\frac{d}{dx}(x(t))=\frac{d}{dx}(sin(t+\pi))\hat{i}+\frac{d}{dx}(sin(2t-\pi/4))\hat{j}=v(t)`
So, `v(t)=cos(t+\pi)\hat{i}+2cos(2t-\pi/4)\hat{j}`

At `t=(-\pi)/3`, `v((-\pi)/3)=cos(-\pi/3+\pi)\hat{i}+2cos(-2\pi/3-\pi/4)\hat{j}=cos(2\pi/3)\hat{i}+2cos(\frac{-8\pi-3\pi}{12\pi})\hat{j}=-0.5\hat{i}--1.931\hat{j}`
So, velocity of the object at time `t=(-\pi)/3 " is " -0.5\hat{i}--1.931\hat{j}`
Try to find the magnitude as an exercise.