Answer to Question #130108 in Mechanics | Relativity for Ariyo Emmanuel

Question

1. The motion of a particle is defined by the relation

Answer 1

The motion of a particle is defined by the relation $x=6t^{4}-2t^{3}-12t^{2}+3t+3$, where x and t expressed in meters and seconds. Determine the time, the position and the velocity when a=0

$v=\frac{dx}{dt}(6t^{4}-2t^{3}-12t^{2}+3t+3)$

$=(4\times6)t^{3}-(3\times2)t^{2}-(12\times2)t+3$

$=24t^{3}-6t^{2}-24t+3 (1)$

$a=\frac{d}{dt}(v)=(24\times3)t^{2}-(2\times6)t-24$

$=72t^{2}-12t-24$

so $0=72t^{2}-12t-24$

$6t^{2}-t-2=0$

$t_{1}=-\frac{1}{2}, t_{2}=\frac{2}{3}$

taking the positive time as correct we substitute to (1)

so $v=24(\frac{2}{3})^{3}-6(\frac{2}{3})^{2}-24(\frac{2}{3})+3$

$=24(0.2962)-6(0.4444)-16+3$

$=7.1088-2.664-16+3=-8.5552 m/sec$

Also $x-(6\times0.1975)-(2\times0.2962)-(12\times0.4444)+2+3$

$=1.185-0.5924-5.3328+2+3$

$=0.2598 m$