Answer
The plane travels a distance of 4.22 km
Work Step by Step
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
