Question

The equation of motion of a particle is given by $s=2t^3–9t^2+12t+1$, where s and t are measured in cm and sec. The time when the particle stops momentarily is

• A. $1sec$
• B. $2sec$
• C. $1,2sec$
• D. None of the above

Answer 1

The correct option is C

$1,2sec$

Step 1: Given Data:

The equation of motion, $s=2t^3–9t^2+12t+1·····\left(1\right)$

Step 2: Formula used:

velocity is defined as the rate of change of displacement

$v=\frac{ds}{dt}$

Step 3: Calculation of time when the particle stops momentarily :

The particle stops momentarily means its velocity becomes zero so

first, find velocity,

$v=\frac{ds}{dt}$

Differentiate equation (1) with respect to time to get velocity-

$$\begin{gathered} v=\frac{d\left(2t^3-9t^2+12t+1\right)}{dt} \\ v=6t^2-18t+12 \\ 0=6t^2-18t+12 \end{gathered}$$

Further simplifying,

$$\begin{gathered} 6t^2-18t+12=0 \\ {t}^2-3t+2=0 \\ {t}^2-2t-t+2=0 \\ t\left(t-2\right)-1\left(t-2\right)=0 \\ \left(t-2\right)\left(t-1\right)=0 \\ t=1,2sec \end{gathered}$$

Hence, option D is the correct answer.