Question

A particle moves along a straight line $OX$. At a time $ts$, the distance $xm$ of the particle from $O$ is given by $x=40+12t–t^3$. How long would the particle travel before coming to rest?

• A. $24m$
• B. $40m$
• C. $56m$
• D. $16m$

Answer 1

The correct option is C

$56m$

Step 1: Given data

The distance travelled by the particle, $x=40+12t–t^3$

Step 2: Find the distance travelled by the particle before coming to rest

Speed (v) is rate of change of distance and given by

$\begin{array}{rcl}v& =& \frac{dx}{dt}\\ v& =& 12-3t^2\\ 0& =& 12-3t^2\\ t& =& 2s\end{array}$ [$v=0$ as body is at rest]

Now, putting thus value in $x$

$$\begin{gathered} x=40+12t-t^3 \\ x=40+12\times 2-{\left(2\right)}^3 \\ x=56m \end{gathered}$$

Therefore, the distance travelled by the particle before coming to rest is $56m$.

Hence, option $\left(C\right)$ is correct.