Integrate over [0, 1] to get the position of the body after 1 second. It starts at the origin, so we take x(0 s) = 0 m and using the fundamental theorem of calculus,
[tex]\displaystyle x(1\,\mathrm s) = x(0\,\mathrm s) + \int_{0\,\rm s}^{1\,\rm s} \left(24\frac{\rm m}{\rm s}\right) \, dt = 24 \, \mathrm m[/tex]
Then for time t ≥ 1 second, we integrate over [1, 11] to get the position after the next 10 seconds of motion. Since x(1 s) = 24 m, we have
[tex]\displaystyle x(11\,\mathrm s) = x(1\,\mathrm s) + \int_{1\,\rm s}^{11\,\rm s} \left(24\frac{\rm m}{\rm s} - \left(6\frac{\rm m}{\mathrm s^2}\right) t\right) \, dt = \boxed{-36\,\mathrm m}[/tex]
Without using calculus, in the 1st second of motion the body travels at constant speed, so it reaches a new position of
x = (24 m/s) (1 s) = 24 m
In the next 10 seconds, the body has constant acceleration and its velocity at time t is
v = 24 m/s - (6 m/s²) (t - 1 s)
so its "initial" velocity is 24 m/s, and after 10 seconds it attains a "final" velocity of
24 m/s - (6 m/s²) (11 s - 1 s) = -36 m/s
Then the body travels a distance x such that
(-36 m/s)² - (24 m/s)² = 2 (-6 m/s²) x ⇒ x = - 60 m
and so the body's net change in position is 24 m - 60 m = -36 m.
