TS EAMCET 5 May 2018 Shift 2 Solved Paper

The velocity attained by the particle at the end of the time is calculated as, $v_{1}=u+at_{1}$ $=4t_{1}$ Therefore, the displacement in first $t_{1}$ seconds is given by, $s_{1}=\frac{1}{2}at_{1}^{2}$ $=\frac{1}{2}\times4\times t_{1}^{2}$ $=2t_{1}^{2}$ The displacement in the next $t_{2}$ seconds is given by, $s_{2}=v_{1}t_{2}$ $=4t_{1}t_{2}$ For third part of journey, Initial velocity is $v_{1}=4t_{1},$ final velocity is 0 and time interval $t_{1}$ Thus, acceleration is given by, $a=\frac{0-4t_{1}}{t_{1}}$ $=-4\ \mathrm{ms}^{-2}$ Thus, displacement in the third part is, $s_{3}=\frac{-16t_{1}^{2}}{-8}$ $=2t_{1}^{2}$ Therefore, $v_{2}^{2}-v_{1}^{2}=2as_{3}$ Now, average velocity is calculated as, $v_{\text{avg }}=\frac{\text{Total displacement }}{\text{Total time }}$ $5.1=\frac{s_{1}+s_{2}+s_{3}}{t_{1}+t_{2}+t_{1}}$ $5.1=\frac{2t_{1}^{2}+4t_{1}t_{2}+2t_{1}^{2}}{10}$ $51=4t_{1}^{2}+4t_{1}(10-2t_{1})$ Solve further. $4t_{1}^{2}+40t_{1}+51=0$ $(2t_{1}-3)(2t_{1}-17)=0$ $t_{1}=\frac{3}{2}$ or $t_{1}=\frac{17}{2}$ $t_{1}=1.5\ \mathrm{s}$ As, $t_{1}=8.5\ \mathrm{s}$ is not possible. It exceeds total time.