Question

The velocity of a particle which starts from rest is given by the following table.

t (in second)	0	2	4	6	8	10
v (in m/s)	0	12	16	20	35	60

The total distance travelled (in metre) by the particles in $10s$, using Trapezoidal rule is given by

• A. 113
• B. 226
• C. 143
• D. 246

Answer 1

Answer: (B)

Given table is

t	0	2	4	6	8	10
v	0	12	16	20	35	60

Here, $h=\frac{10-0}{5}=2$
$\therefore$ Total distance $=\frac{h}{2}[f\left(x_0\right)+2\{f\left(x_1\right)+f\left(x_2\right)$
$+f\left(x_3\right)+f\left(x_4\right)\}+f\left(x_5\right)]$
$=\frac{2}{2}[0+2(12+16+20+35)+60]$
$=166+60=226$