$\text{v}(\text{t})=2\text{t}(3-\text{t})=6\text{t}-2\text{t}^2$
Distance covered in 0 to 3s = 9m Distance covered in 3 to 6s$=\int_3^6(18-9\text{t}+\text{t}^2)\text{dt}=\Big(18\text{t}-\frac{9\text{t}^2}{2}+\frac{\text{t}^3}{3}\Big)^6$
$=18\times6-\frac{9}{2}\times6^2+\frac{6^3}{3}-\Big(18\times3-\frac{9\times3^2}{2}+\frac{3^3}{3}\Big)$
$=108-9\times18+\frac{6^3}{3}-18\times3+\frac{9}{2}\times9-\frac{27}{3}$
$=-4.5\text{m}$
$\therefore$ Total distance travelled in one cycle
$=\text{s}_1+\text{s}_2=9-4.5=4.5\text{m}$
Number of cycles to be covered in total distance $=\frac{20}{4.5}\approx4.44\approx5$Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
$\text{f}=-\frac{1}{2\pi}\bigg(\frac{\text{v}^2_1-\text{v}^2_2}{\text{x}^2_2-\text{x}^2_1}\bigg)^{\frac{1}{2}}$