Procedure
Part A
a ) v = vo + a*t ( the acceleration will be negative )
9.50 m/s = 16 m/s + a*1.5
[tex]\begin{gathered} a=\frac{9.50-16}{1.5} \\ a=-4.33m/s^2 \end{gathered}[/tex]Now, the force
[tex]\begin{gathered} F=ma \\ F=-980\operatorname{kg}\cdot4.33m/s^2 \\ F=-4243.4N \end{gathered}[/tex]Part B
Opposite to the direction of motion
Part C
[tex]\begin{gathered} d=v_ot-\frac{at^2}{2} \\ d=16\cdot1.5-\frac{4.33\cdot1.5^2}{2} \\ d=19.12\text{ m} \end{gathered}[/tex]