The position of a particle moving along the x | vedclass

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Question

(C) Given the position function: $x = -2t^3 + 3t^2 + 5$.Velocity $v$ is the first derivative of position with respect to time: $v = \frac{dx}{dt} = \f

Vedclass · Accepted Answer

$-6$

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(C) Given the position function: $x = -2t^3 + 3t^2 + 5$.Velocity $v$ is the first derivative of position with respect to time: $v = \frac{dx}{dt} = \frac{d}{dt}(-2t^3 + 3t^2 + 5) = -6t^2 + 6t$.Set the velocity to zero to find the time $t$: $-6t^2 + 6t = 0 \Rightarrow 6t(1 - t) = 0$. This gives $t = 0 \ s$ or $t = 1 \ s$. Since the particle is moving,we consider $t = 1 \ s$.Acceleration $a$ is the derivative of velocity with respect to time: $a = \frac{dv}{dt} = \frac{d}{dt}(-6t^2 + 6t) = -12t + 6$.Substitute $t = 1 \ s$ into the acceleration equation: $a = -12(1) + 6 = -6 \ m/s^2$.

Column-$I$	Column-$II$
$(1)$ Velocity-time relation	$(a)$ $v = v_0 + at$
$(2)$ Velocity-displacement relation	$(b)$ $S = v_0t + \frac{1}{2}at^2$
	$(c)$ $v^2 = v_0^2 + 2as$

The position of a particle moving along the $x$-axis is given by $x = (-2t^3 + 3t^2 + 5) \ m$. The acceleration of the particle at the instant its velocity becomes zero is ....... $m/s^2$.

Similar Questions

Match the relations in Column-$I$ with the equations in Column-$II$ for uniformly accelerated motion.

Column-$I$ Column-$II$

$(1)$ Velocity-time relation $(a)$ $v = v_0 + at$

$(2)$ Velocity-displacement relation $(b)$ $S = v_0t + \frac{1}{2}at^2$

$(c)$ $v^2 = v_0^2 + 2as$

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