Question

A particle moves along x-axis in such a way that its $x$-co-ordinate varies with time according to the equation $x = 8 - 4t + 6t^2$. The distance covered by particle between $t =0$ to $t =\frac{2}{3}$ sec . is:

• A. Zero
• B. $\frac{8}{3} m$
• C. $8 m$
• D. $\frac{4}{3} m$

Answer 1

Answer: (D)

$X_{t=0}=8, \frac{d x}{d t}=0$
$\Rightarrow t=\frac{1}{3}$
$X_{t=\frac{1}{3}}=8-\frac{4}{3}+\frac{6}{9}$
$=\frac{22}{3}$
$X_{t=2 / 3}=8$
distance between $t=0$ to $t=\frac{2}{3}$ is
$2\left(8-\frac{22}{3}\right)=\frac{4}{3} m$