Tracking a Particle's Distance

Since $a=\frac{d^2s}{dt^2}=\frac{dv}{dt}=2t,$ it follows that $v(t)=\frac{ds}{dt}=\int 2t dt=t^2+C_1,$ where $C_1$ is the constant of integration.

Since $v(1)=6$ yields $C_1=5,$ thus $v(t)=t^2+5.$ Since $v(t)=t^2+5 \geq 0$ for all $t \geq 0,$ the distance traveled is $s(t)=\int (t^2+5) dt=\frac{1}{3}t^3+5t+C_2,$ where $C_2$ is the constant of integration.

Since $s(1)=17$ yields $C_2=\frac{35}{3},$ thus, $s(t)=\frac{1}{3}t^3+5t+\frac{35}{3}.$ Therefore, $s(2)=\frac{1}{3} \cdot 2^3+5 \cdot 2+\frac{35}{3}=\frac{73}{3}.$