# What is the difference between t = 0 and t = 1 seconds?

Aug 12, 2018

Given displacement vector

$x = 4 \cos \left(\pi t + \frac{\pi}{4}\right)$

Value of $x$ at $t = 0$

$x \left(0\right) = 4 \cos \left(\pi \times 0 + \frac{\pi}{4}\right)$
$\implies x \left(0\right) = 4 \cos \left(\frac{\pi}{4}\right)$

Value of $x$ at $t = 1$

$x \left(1\right) = 4 \cos \left(\pi \times 1 + \frac{\pi}{4}\right)$
$\implies x \left(1\right) = 4 \cos \left(\frac{5 \pi}{4}\right)$

Displacement between $t = 0 \mathmr{and} t = 1$ is

$\Delta x = x \left(1\right) - x \left(0\right)$
$\implies \Delta x = 4 \cos \left(\frac{5 \pi}{4}\right) - 4 \cos \left(\frac{5 \pi}{4}\right)$
$\implies \Delta x = 4 \left[\cos \left(\frac{5 \pi}{4}\right) - \cos \left(\frac{\pi}{4}\right)\right]$

Making use of the unit circle reproduced below to rewrite in vector form. The first value in the round brackets represents $\cos$ or $\hat{j}$ and second value represents $\sin$ or $\hat{i}$ component for a particular angle.

$\implies \Delta x = 4 \left[- \frac{\sqrt{2}}{2} \hat{j} - \frac{\sqrt{2}}{2} \hat{j}\right]$
$\implies \Delta x = - 4 \sqrt{2} \hat{j}$

Aug 12, 2018

This is 1-D motion, along the number line:

• $x \left(t\right) = 4 \cos \left(\pi t + \frac{\pi}{4}\right)$

• $\left\{\begin{matrix}x \left(0\right) = 4 \cos \left(\frac{\pi}{4}\right) = \frac{4}{\sqrt{2}} \\ x \left(1\right) = 4 \cos \left(\frac{5 \pi}{4}\right) = - \frac{4}{\sqrt{2}}\end{matrix}\right.$

$\Delta x = - \frac{4}{\sqrt{2}} - \frac{4}{\sqrt{2}} = - \frac{8}{\sqrt{2}} = - 4 \sqrt{2} \text{ metres }$