Question

A water-skier is moving at a speed of 14.3 m/s. When she skis in the same direction as a traveling wave, she springs upward every 0.450 s because of the wave crests. When she skis in the direction opposite to that in which the wave moves, she springs upward every 0.342 s in response to the crests. The speed of the skier is greater than the speed of the wave. Determine (a) the speed and (b) the wavelength of the wave.

Answer 1

Answer:

a) 1.95 m/s

b) 5.56 m

Explanation:

Given that:

Velocity of the skier $(V_s)$ = 14.3 m/s

For the skier moving in the direction of the wave, we have:

Period (T) = 0.450 s

Relative velocity (V) of the skier in regard with the wave =  $(V_s - V_w)$

where:

$V_s$ = velocity of the skier

$V_w$ = velocity of the wave

The wavelength $(\lambda)$ can be written as:

$\lambda = (V_s-V_w)T$

$\lambda = (V_s-V_w) 0.450m$ ---------------> Equation (1)

For the skier moving opposite in the direction of the wave, we have:

Period (T) = 0.342 s

Relative velocity (V) of the skier in regard with the wave = $(V_s + V_w)$

The wavelength $(\lambda)$ can be written as:

$\lambda = (V_s+V_w)T$

$\lambda = (V_s+V_w) 0.342m$   ------------------> Equation 2

Equating equation (1) and equation (2) and substituting  $V_s$  = 14.3 m/s ; we have:

$(V_s-V_w) 0.450m = (V_s-V_w) 0.342m$

$0.450m(V_s)-0.450m(V_w) = 0.342m(V_s)+0.342m(V_w)$

Collecting the like terms; we have:

$0.450m(V_s) - 0.342m(V_s) = 0.342m(V_w)+0.450m(V_w)$

$(V_s)(0.450m - 0.342m) = (V_w)0.342m+0.450m$

$14.3m/s(0.450m - 0.342m) = (V_w)0.342m+0.450m$

$14.3m/s(0.108m = (V_w)0.792m$

$1.5444m^2/s = (V_w)0.792m$

$(V_w) = \frac{1.5444m^2/s}{ 0.792m}$

$(V_w) = 1.95 m/s$

b)

The Wavelength of the wave can be calculated using :  $( \lambda }) = (V_s-V_w) 0.450m$

$({\lambda}) = (14.3 m/s -1.95 m/s)(0.450)$

$(\lambda) = (12.35)0.450m$

$(\lambda)= 5.5575 m$

λ ≅ 5.56 m