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.

Column $I$	Column $II$
$(A)$ Distance travelled in $3\,s$	$(p)$ $-20$ units
$(B)$ Displacement in $1\,s$	$(q)$ $15$ units
$(C)$ Initial acceleration	$(r)$ $25$ units
$(D)$ Velocity at $4\,s$	$(s)$ $-10$ units

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In the $s-t$ equation $(s=10+20t-5t^2)$,match the following columns.

Column $I$ Column $II$

$(A)$ Distance travelled in $3\,s$ $(p)$ $-20$ units

$(B)$ Displacement in $1\,s$ $(q)$ $15$ units

$(C)$ Initial acceleration $(r)$ $25$ units

$(D)$ Velocity at $4\,s$ $(s)$ $-10$ units

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