The motion of a particle along a straight lin | vedclass

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Question

$\begin{array}{l}
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,Give\,:\,x = 8 + 12t - {t^3}\
\,velocity\,,\,v = \frac{{dx}}{{dt}} = 12 - 3{t^2}\
When\,v =

Vedclass · Accepted Answer

$12$

Vedclass · Answer

$\begin{array}{l}
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,Give\,:\,x = 8 + 12t - {t^3}\
\,velocity\,,\,v = \frac{{dx}}{{dt}} = 12 - 3{t^2}\
When\,v = 0,\,12 - 3{t^2} = 0\,\,\,or\,\,t = 2\,s\
\,\,\,\,\,\,\,\,\,\,\,a = \frac{{dv}}{{dt}} =  - 6t\
\,\,\,\,\,\,\,\,\,\,\,a\left| {_{t = 2\,s} =  - 12\,m\,{s^{ - 2}}} ight.\
{m{Retardation}}\, = 12\,m\,{s^{ - 2}}
\end{array}$

The motion of a particle along a straight line is described by equation $x = 8 + 12t - t^3$ where $x$ is in metre and $t$ in second. The retardation of the particle when its velocity becomes zero is...........$m/s^2$

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