At time t = 0, a car moving along a straight line has a velocity of 16 m/s.  It slows down with an acceleration of -0.5t m/s², where t is in seconds. Mark the correct statement(s)

This is the example of non-uniform acceleration  $a=\frac{dv}{dt}=-0.5t$
 $\underset{16}{\overset{v}{\int }}dv=-\underset{0}{\overset{t}{\int }}0.5tdt\Rightarrow v=16-\frac{0.5t^2}{2}$
Direction of velocity changes at the moment when it becomes zero monentarily
 $0=16\frac{0.5t^2}{2}\Rightarrow t=8s\frac{dx}{dt}=v=16-\frac{0.5t}{2}$
Let us consider that at t = 0, particle is at x = 0
 $\underset{0}{\overset{x}{\int }}dx=\underset{0}{\overset{t}{\int }}(16-\frac{0.5t^2}{2});x=16t-\frac{0.5t}{6};$
Distance travelled = |displacement| for $t\underset{\_}{<}8s$  . So, distance travelled in 4s
 $x=16\times 4-\frac{0.5\times 4^3}{6}=59m$
Distance travelled in 10s = |Displacement in 8s| x 2-
Displacement in  $10s=85.33\times 2-76.55=94mv(t=10s)=-9m{s}^{-1}$