Position of a particle moving along x-axis is given by x=2+8 t-4 t2. The distance travelled by the particle from t=0 to t=2 is :

Question

Position of a particle moving along x-axis is given by x=2+8 t-4 t2. The distance travelled by the particle from t=0 to t=2 is : (A) 0 (B) 8 (C) 12 (D) 16

Tardigrade Admin · Accepted Answer

Area of ( v - t ) curve = displacements x =2+8 t -4 t 2

v=(d x/d t)=8-8 t

Distance travelled from t =0 to t =2 sec =(1/2) × 1 × 8+(1/2) × 1 × 8=4+4=8 m

Q. Position of a particle moving along $x$ -axis is given by $x = 2 + 8 t - 4 t^{2}$ . The distance travelled by the particle from $t = 0$ to $t = 2$ is :

0

8

12

16

Area of $(v - t)$ curve $=$ displacements
$x = 2 + 8 t - 4 t^{2}$

$v = \frac{d x}{d t} = 8 - 8 t$

Distance travelled from $t = 0$ to $t = 2 sec$
$= \frac{1}{2} \times 1 \times 8 + \frac{1}{2} \times 1 \times 8 = 4 + 4 = 8 m$

Q. Position of a particle moving along x-axis is given by x=2+8t−4t2. The distance travelled by the particle from t=0 to t=2 is :

0

8

12

16

Area of (v−t) curve = displacements x=2+8t−4t2 v=dtdx​=8−8t Distance travelled from t=0 to t=2sec =21​×1×8+21​×1×8=4+4=8m

Q. Position of a particle moving along $x$ -axis is given by $x = 2 + 8 t - 4 t^{2}$ . The distance travelled by the particle from $t = 0$ to $t = 2$ is :

Area of $(v - t)$ curve $=$ displacements
$x = 2 + 8 t - 4 t^{2}$

$v = \frac{d x}{d t} = 8 - 8 t$

Distance travelled from $t = 0$ to $t = 2 sec$
$= \frac{1}{2} \times 1 \times 8 + \frac{1}{2} \times 1 \times 8 = 4 + 4 = 8 m$