• question_answer The cord mates of a moving particle at any time t are given by $X={{t}^{3}}$ and $Y=4{{t}^{2}},$ where X and Y are in meter and t in second. The acceleration of the particle at time $t-1s,$ is given by: A)  $6m{{s}^{-2}}$                B)  $8m{{s}^{-2}}$C)  $10m{{s}^{-2}}$                D)  $14m{{s}^{-2}}$

Given that,     $X={{t}^{3}},\,Y=4{{t}^{2}},Vx=3{{t}^{2}}$ and              ${{a}_{x}}=6t$                 (at  $t=1s$) $=6m/{{s}^{2}}$ ${{V}_{Y}}=8t,{{a}_{Y}}=8m/{{s}^{2}}$ (at  $t=15$)                 $a=\sqrt{a_{X}^{2}+a_{Y}^{2}}$                 $a=\sqrt{{{(6)}^{2}}+{{(8)}^{2}}}=10m/{{s}^{2}}$