(a) Consider that the distance function s(t) can be obtained by integrating the function v(t), as follow:
[tex]s(t)=\int v(t)dt[/tex]
under the condition that s(1) = 50 m.
(b) By integrating you have:
[tex]\begin{gathered} s(t)=\int (6t^2-4t+10)dt \\ s(t)=6\frac{t^3}{3}-4\frac{t^2}{2}+10t+C \\ s(t)=2t^3-2t^2+10t+C \end{gathered}[/tex]
use the IVP s(1) = 50 and solve for C:
[tex]\begin{gathered} s(1)=2(1)^3-2(1)^2+10(1)+C=50 \\ s(1)=2-2+10+C=50 \\ 10+C=50 \\ C=40 \end{gathered}[/tex]
Then, the function s(t) is:
[tex]s(t)=2t^3-2t^2+10t+40[/tex]
(c) For t=10s you have:
[tex]\begin{gathered} s(10)=2(10)^3-2(10)^2+10(10)+40 \\ s(10)=2000-200+100+40 \\ s(10)=1940 \end{gathered}[/tex]
The distance traveled in 10 s is 1940 m.
(d) The acceleration is the first derivative of v(t), then, you have:
[tex]\begin{gathered} a(t)=v(t)^{\prime}=12t-4 \\ a(5)=12(5)-4=60-4=56 \end{gathered}[/tex]
The acceleration is 56m/s^2
