Question

The velocity of a particle at time $t$ is given by the relation $v=6t-\left(\frac{t^2}{6}\right)$. The distance travelled in $3sec$ is, if $s=0$at $t=0$

• A. $\raisebox{1ex}{$39$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$
• B. $\raisebox{1ex}{$57$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$
• C. $\raisebox{1ex}{$51$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$
• D. $\raisebox{1ex}{$33$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$

Answer 1

The correct option is C

$\raisebox{1ex}{$51$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$

Step 1: Given data:

The velocity of a particle at time $t$ is given by the relation,

$v=6t-\left(\frac{t^2}{6}\right)\dots \dots \dots \dots \dots \dots \dots \dots \left(1\right)$

Time taken $t=3sec$

At $t=0$, the distance travelled $s=0$

Step 2: Formula used:

Velocity is defined as the rate of change of displacement. that is,

$v=\frac{ds}{dt}\dots \dots \dots \dots \dots \dots \dots \left(2\right)$

where, $v,s$ are velocity and distance travelled respectively.

Step 3: Calculation of displacement equation:

We know from the equation (2),

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

It can be written as-

$ds=vdt$

Put the value from equation (1), and integrating both the sides

$\begin{array}{rcl}\int ds& =& \int \left(6t-\frac{t^2}{6}\right)dt\\ s& =& 6\left(\frac{t^2}{2}\right)-\frac{1}{6}\left(\frac{t^3}{3}\right)+c\\ s& =& 3t^2-\frac{t^3}{18}+c\dots \dots \dots \dots \dots \dots \dots \left(3\right)\end{array}$$c=cons\tan t$

putting the given condition, at $t=0$, $s=0$ in equation (3) so we get value of constant as,

$$\begin{gathered} 0=0-0+c \\ c=0 \end{gathered}$$

putting the value of the constant in equation (3)

Thus the displacement equation will be-

$s=3t^2-\frac{t^3}{18}\dots \dots \dots \dots \left(4\right)$

Step 4: Calculation of distance travelled in $t=3sec$:

Now, put the value of $t=3sec$ in equation (4) we get-

The distance can be calculated as-

$$\begin{gathered} s=3\left(3^2\right)-\frac{3^3}{18} \\ s=27-\frac{27}{18} \\ s=27-\frac{3}{2} \\ s=27-1.5 \\ s=25.5=\frac{51}{2}unit \end{gathered}$$

Thus, the distance travelled in $3sec$ is $\raisebox{1ex}{$51$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$.

Hence, option (C) is the correct answer.