A particle moves along a straight line $\mathrm{OX}$. At a time $t$ (in seconds) the distance $x$ (in metres) of the particle from $\mathrm{O}$ is given by $x=40+12 t-t^3$. How long would the particle travel before coming to rest ?

Question

Aicharya · Accepted Answer

The given equation is
$x=40+127-t^3$
Differentiating we get
$u=\frac{d x}{d t}=12-3 t^2$
When particle comes to rest means
$\begin{array}{l}
& v = 0 \\
& 12-3 t^2 =0 \\
& t =2 \mathrm{sec} .
\end{array}$
The distance travelled before coming to rest
$\begin{aligned}
\int_0^5 d s & ={ }_0^2 v d t \\
5 & =\int_0^2\left(12-3 t^2\right) d t \\
& =\left[12 t-\frac{3 t^3}{3}\right]_0^2 \\
& =12 \times 2-(8-0) \\
& =16 \mathrm{~m} .
\end{aligned}$

A particle moves along a straight line $\mathrm{OX}$. At a time $t$ (in seconds) the distance $x$ (in…

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