A particle starts moving from time t=0 and it | vedclass

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(B) Given position: $x(t) = 4t^3 - 3t$.$1$. To find when it returns to the origin,set $x(t) = 0$: $t(4t^2 - 3) = 0$. Since $t > 0$,$t = \sqrt{3/4} = \

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$A, B, C$ only

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(B) Given position: $x(t) = 4t^3 - 3t$.$1$. To find when it returns to the origin,set $x(t) = 0$: $t(4t^2 - 3) = 0$. Since $t > 0$,$t = \sqrt{3/4} = \frac{\sqrt{3}}{2} \approx 0.866$. Statement $A$ is correct.$2$. Velocity $v(t) = \frac{dx}{dt} = 12t^2 - 3$. At the turning point,$v = 0$,so $12t^2 = 3 \Rightarrow t^2 = 1/4 \Rightarrow t = 0.5$.$3$. Position at turning point: $x(0.5) = 4(0.5)^3 - 3(0.5) = 4(0.125) - 1.5 = 0.5 - 1.5 = -1$. The particle is $|-1| = 1$ unit away from the origin. Statement $B$ is correct.$4$. Acceleration $a(t) = \frac{dv}{dt} = 24t$. For $t \ge 0$,$a(t) \ge 0$. Statement $C$ is correct.$5$. Since $a(t) \ge 0$ for $t > 0$,the velocity increases,but the particle can still turn back if it starts with negative velocity. Here,$v(0) = -3$,so it moves in the negative direction until $t=0.5$,then turns back. Statement $E$ is incorrect.Thus,statements $A, B, C$ are correct.

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