Hence \(\omega = \)coefficient of \(t = 2 \pi\)
==> Maximum speed of the particle \({v_{\max }} = a\omega = 10 \times 2\pi \)
\(= 10 × 2 × 3.14 = 62.8 ≈ 63 cm/s\)
$(a)$ $\left(x^2-v t\right)^2$
$(b)$ $\log \left[\frac{(x+v t)}{x_0}\right]$
$(c)$ $e^{\left\{-\frac{(x+v t)}{x_0}\right\}^2}$
$(d)$ $\frac{1}{x+v t}$