The position(x) of a particle at any time(t) | vedclass

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Question

$\begin{array}{l}
We\,have\,X\left( t ight) = 4{t^3} - 3{t^2} + 2\
 \Rightarrow \,V = \,\frac{{dv}}{{dt}} = 12{t^2} - 6t\
and\,a\, = \frac

Vedclass · Accepted Answer

$42\, ms^{-2}$ and $36\, ms^{-1}$

Vedclass · Answer

$\begin{array}{l}
We\,have\,X\left( t ight) = 4{t^3} - 3{t^2} + 2\
 \Rightarrow \,V = \,\frac{{dv}}{{dt}} = 12{t^2} - 6t\
and\,a\, = \frac{{dv}}{{dt}} = 24t\
	herefore \,V\,at\,\,t = 2s\,is\,12{\left( 2 ight)^2} - 6\left( 2 ight)\
i.e.,36m{s^{ - 1}}\
and\,a\,at\,t\, = 2s\,is\,24 	imes \,2i.e.,46m{s^{ - 2}}
\end{array}$

The position$(x)$ of a particle at any time$(t)$ is given by $x(t) = 4t^3 -3t^2 + 2$ The acceleration and velocity of the particle at any time $t = 2\, sec$ are respectively

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