The velocity of a particle which starts from rest is given by the following table.
The total distance travelled (in metre) by the particles in $10 \mathrm{~s}$, using Trapezoidal rule is given by

• A

  113
• Correct answer
  B

  226
• C

  143
• D

  246

Solution

Given table is
Here, $\quad h=\frac{10-0}{5}=2$ $\begin{aligned} & \therefore \text { Total distance }=\frac{h}{2}\left[f\left(x_0\right)+2\left\{f\left(x_1\right)+f\left(x_2\right)\right.\right. \\ & \left.\left.+f\left(x_3\right)+f\left(x_4\right)\right\}+f\left(x_5\right)\right] \\ & =\frac{2}{2}[0+2(12+16+20+35)+60] \\ & =166+60=226 \\ & \end{aligned}$

Hence, the correct answer is option: 'B'.