Answer:
a)A= 3 m
b)[tex]y(x)=\dfrac{6}{(x-3t)^2+2}\ m[/tex]
c)D= 15 m
Explanation:
Given that
[tex]y(x)=\dfrac{6}{x^2+2}\ m[/tex]
v= 3 m/s
a)
The amplitude(A) of the pulse :
When x= 0 ,Then y = A
Put x= 0
[tex]y(x)=\dfrac{6}{x^2+2}\ m[/tex]
[tex]y(0)=\dfrac{6}{0^2+2}\ m[/tex]
y= A= 3 m
A= 3 m
b)
Distance travel in time t
x= vt
x= 3 t
[tex]y(x)=\dfrac{6}{(x-3t)^2+2}\ m[/tex]
c)
The distance covered by pulse in the time 5 s
D = v t
D= 3 x 5
D= 15 m
