A particle moves along a straight line and pusses through a fixed point O with a velocity of [math]12ms^{-1}[/math]. The acceleration, a [math]ms^{-2}[/math], at t seconds after passing through O is given by [math]a=4-2t[/math] [br](a) Determine the instantaneous displacement, in me, of the particle from O[br](i) when [math]t=3[/math] (ii) when the particle is at rest[br](b) Next, find distance, in m, travelled by the particle in the 7th second[br]

[b]Solution[/b][br][br]Velocity function, v is given by [math]v=\int a^{^{_{_{ }}}}dt[/math][br] [math]v=\int(4-2x)dt[/math][br] [math]v=4t-t^2+c[/math][br]When [math]t=0[/math] and [math]v=12[/math], then [math]12=4(0)-(0)^2+c[/math][br] [math]c=12[/math][br]Hence, at time t, [math]v=12+4t-t^2[/math][br]Displacement function, a is given by,[math]x=\int vdt[/math][br] [math]s=\int(12+4t-t^2)dt[/math][br][math]s=12x+2t^2-1/3t^3+c[/math][br]When [math]t=0[/math] and [math]s=0[/math], then, [math]0=12(0)+2(0)^2-1/3(0)^3+c[/math][br] [math]c=0[/math][br]Hence, at time t, [math]s=12t+2t^2-1/3t[/math][br](a) (i) When[math]t=3[/math][i], [/i][math]s=12(3)+2(3)^2-1/3(3)^3[/math][math]s=36+18-9[/math][math]s=45[/math][br]Hence, the instantaneous displacement when[math]t=3[/math] is 45 m.[br][br](ii) When the particle is at rest, [math]v=0[/math]. [br]Then, [math]12+4t-t^2=0[/math][math](t+2)(t-6)=0[/math] Since [math]t\ge0[/math][i], [/i][math]t=6[/math][br]When [math]t=6[/math][i], [/i][math]s=12(6)+2(6)^2-1/3(6)^3[/math][math]s=72+72-72[/math][math]s=72[/math][br]Hence, the instantaneous displacement when the particle is at rest is 72 m.[br][br](b) When [math]t=7,_{^{_{ }}}^{^{^{_{_{_{_{_{_{_{ }}}}}}}}}}s=12(7)+2(7)^2-\frac{1}{3}(7)^3[/math][br] [math]s=84+98-\frac{1141}{3}[/math] [br] [math]s=\frac{203}{3}[/math][br][br] From the number the distance travelled by the particle in the 7th second [br]    [math]=|s7-s6|[/math][br] [math]=|\frac{203}{3}-72|[/math][br] [math]=\frac{13}{3}[/math]