Answer with Explanation:
The position of the particle as a function of time is given by [tex]s(t)=t^2-26t+133[/tex]
Part 1) The position as a function of time is shown in the below attached figure.
Part 2) By the definition of velocity we have
[tex]v=\frac{ds}{dt}\\\\\therefore v(t)=\frac{d}{dt}\cdot (t^2-26t+133)\\\\v(t)=2t-26[/tex]
The velocity as a function of time is shown in the below attached figure.
Part 3) The displacement of the particle in the first 19 seconds is given by [tex]\Delta x=s(19)-s(0)\\\\\Delta x=(19^2-26\times 19+133)-(0-0+133)=-133millimeters[/tex]
Part 4) The distance covered in the first 19 seconds can be found by evaluating the integral
[tex]s=\int _{0}^{19}\sqrt{1+(\frac{ds}{dt})^2}\\\\s=\int _{0}^{19}\sqrt{1+(2t-26)^2}\\\\\therefore s=207.03meters[/tex]
Part 4) As we can see that the position-time graph is parabolic in shape hence we conclude that the motion is uniformly accelerated motion.
