Question

The coordinates of a moving particle at any time ${}^{\prime }{t}^{\prime }$ are given by $x=t^3$ and $y=4t^2$, where $x$ and $y$ are in metre and $t$ in second. The acceleration of the particle at time $t=1s$ is given by

Answer 1

$x=t^3,v_x=3t^2,a_x=6t$
$\therefore a_x(t=1s)=6{\text{m/s}}^2$

$y=4t^2,v_y=8t,a_y=8{\text{m/s}}^2$,
$a=\sqrt{a_x^2+a_y^2}=\sqrt{6^2+8^2},a=10{\text{m/s}}^2$
$\therefore$ (c)