## College Physics (4th Edition)

(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 $~t = 1.67~s$
(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 $~t = 1.67~s$