given that initial velocity is
[tex]v_i = 11 m/s[/tex]
deceleration is given as
[tex]a = -0.3 m/s^2[/tex]
now we have to find the distance covered in 16 s
[tex]d = v_i * t + \frac{1}{2} at^2[/tex]
[tex]d = 11*16 - \frac{1}{2}*0.3*16^2[/tex]
[tex]d = 137.6 m[/tex]
so it will cover 137.6 m distance
part b)
in order to find the final speed
[tex]v_f = v_i + at[/tex]
[tex]v_f = 11 - 0.3*16[/tex]
[tex]v_f = 6.2 m/s[/tex]
so its speed will be 6.2 m/s