Initial velocity, $u = 0,\ a = 1\ \mathrm{m\,s}^{-2},\ t = 10\ \mathrm{s}$ , If $S_n$ be the distance covered by the three-wheeler in nth second. Then $S_{n}=u+\frac{a}{2}(2n-1),$ $S_{n}=\frac{a}{2}(2n-1)=\frac{1}{2}(2n-1)$For $n=1,$ $S_{1}=\frac{1}{2}(2\times 1-1)=0.5\ \mathrm{m}$ $n=2,$ $S_{2}=\frac{1}{2}(2\times 2-1)=1.5\ \mathrm{m}$ $n=3,$ $S_{3}=\frac{1}{2}(2\times 3-1)=2.5\ \mathrm{m}$ and so onThe following is the table for the distance $S_n^{\text{th}}$ travelled in $n_{\text{th}}$ second.Also $D_n=$ distance covered by the particle in 10 sec.$D_{n}=u n+\frac{1}{2}a n^{2}$$=0\times 10+\frac{1}{2}\times 1\times 10^{2}$$(n=10)$$= 50\ \mathrm{m}.$As the equation describing the relation between $D_n$ and n is of seconddegree, so the graph of the vehicle from start to 10$^{\text{th}}$second is expectedto be a parabola. Velocity of the vehicle at the end of 10$^{\text{th}}$ second is$v = 0 + 1 \times 10 = 10\ \mathrm{m\,s}^{-1}$ and it will move with this velocity after 10 seconds.Thus the graph will be a straight line inclined to time axis for uniformlyaccelerated motion. The plot of the distance covered by the three wheelerwith time is shown in the graph.