## College Physics (4th Edition)

Note that the sum of the vertical forces is zero. Let east be the positive direction. We can find the net horizontal force: $\sum F_x = 1.800~kN-1.400~kN = 400~N$ We can find the acceleration: $\sum F_x = m~a_x$ $a_x = \frac{\sum F_x}{m}$ $a_x = \frac{400~N}{1160~kg}$ $a_x = 0.345~m/s^2$ We can find the distance the plane travels in the next $60.0~s$: $\Delta x = v_{0x}~t+\frac{1}{2}a_xt^2$ $\Delta x = (60.0~m/s)(60.0~s)+\frac{1}{2}(0.345~m/s^2)(60.0~s)^2$ $\Delta x = 4220~m = 4.22~km$ The plane travels a distance of 4.22 km