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 + 6t2. The distance covered by particle between t =0 to t =(2/3) sec . is:

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 + 6t2. The distance covered by particle between t =0 to t =(2/3) sec . is: (A) Zero (B) (8/3) m (C) 8 m (D) (4/3) m

Tardigrade Admin · Accepted Answer

Xt=0=8, (d x/d t)=0 ⇒ t=(1/3) Xt=(1/3)=8-(4/3)+(6/9) =(22/3) Xt=2 / 3=8 distance between t=0 to t=(2/3) is 2(8-(22/3))=(4/3) m

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

Zero

$\frac{8}{3} m$

$8 m$

$\frac{4}{3} m$

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

Q. 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+6t2. The distance covered by particle between t=0 to t=32​ sec . is:

Zero

38​m

8m

34​m

Xt=0​=8,dtdx​=0 ⇒t=31​ Xt=31​​=8−34​+96​ =322​ Xt=2/3​=8 distance between t=0 to t=32​ is 2(8−322​)=34​m

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

$\frac{8}{3} m$

$8 m$

$\frac{4}{3} m$

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