#### Answer

(a) Wave I has the fastest speed of $1.50~cm/s$
(b) Wave II has the longest wavelength with a wavelength of $2.09~cm$
(c) Wave II has the fastest maximum speed of a point in the medium with a speed of $13.5~cm/s$
(d) Wave II is moving in the positive x-direction.

#### Work Step by Step

In general: $y(x,t) = A~sin(k~x+\omega~t)$
$I. ~y(x,t) = (1.50~cm)~sin~[~(4.00~cm^{-1})~x+(6.00~s^{-1})~t)~]$
$II. ~y(x,t) = (4.50~cm)~sin~[~(3.00~cm^{-1})~x-(3.00~s^{-1})~t)~]$
(a) $v_1 = \frac{\omega}{k} = \frac{6.00~s^{-1}}{4.00~cm^{-1}} = 1.50~cm/s$
$v_2 = \frac{\omega}{k} = \frac{3.00~s^{-1}}{3.00~cm^{-1}} = 1.00~cm/s$
Wave I has the fastest speed of $1.50~cm/s$
(b) $\lambda_1 = \frac{2\pi}{k} = \frac{2\pi}{4.00~cm^{-1}} = 1.57~cm$
$\lambda_2 = \frac{2\pi}{k} = \frac{2\pi}{3.00~cm^{-1}} = 2.09~cm$
Wave II has the longest wavelength with a wavelength of $2.09~cm$
(c) We can find the maximum speed of wave I:
$v_m = A~\omega$
$v_m = (1.50~cm)(6.00~s^{-1})$
$v_m = 9.00~cm/s$
We can find the maximum speed of wave II:
$v_m = A~\omega$
$v_m = (4.50~cm)(3.00~s^{-1})$
$v_m = 13.5~cm/s$
Wave II has the fastest maximum speed of a point in the medium with a speed of $13.5~cm/s$
(d) The term $(kx-\omega~t)$ shows that a wave is moving in the positive x-direction. Since wave II has the term $(3.00~cm^{-1})~x-(3.00~s^{-1})~t$, we can see that wave II is moving in the positive x-direction.