Answer
(a) At $t = 3.00~s$, the position of the peak is $6.00~m$
(b) The peak of the pulse arrives at $~x = 4.00~m~$ at $~t = 1.67~s$
Work Step by Step
(a) We can find the speed of the wave:
$v = \frac{\Delta x}{\Delta t}$
$v = \frac{1.80~m-1.50~m}{0.20~s}$
$v = 1.50~m/s$
We can find the position at $t = 3.00~s$:
$x = x_0+v~t$
$x = (1.50~m)+(1.50~m/s)(3.00~s)$
$x = 6.00~m$
At $t = 3.00~s$, the position of the peak is $6.00~m$
(b) We can find the time when the position of the peak is $x = 4.00~m$:
$x = x_0+v~t$
$ t = \frac{x - x_0}{v}$
$ t = \frac{4.00~m - 1.50~m}{1.50~m/s}$
$t = 1.67~s$
The peak of the pulse arrives at $~x = 4.00~m~$ at $~t = 1.67~s$
