Question

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

Answer 1

Answer:

The distance traveled by the particle is 8 m.

Explanation:

Given that,

Position of the particle $x=2+8t-4t^2$.....(I)

The distance traveled by the particle from t = 0 to t = 2.

The velocity of the particle is

$v =\dfrac{dx}{dt}$

$v=\dfrac{d}{dt}{(2+8t-4t^2)}$

$v(t)=8-8t$

At t = 1, the velocity is zero

$v(1)=0$

It means particle stop and moves in reverse direction.

Area under the curve of velocity-time graph gives displacement of particle.

Distance = Absolute of v(t)

$D=\int_0^2|v(t)|dt$

$D=\int_{0}^{1}(8-8t)dt-\int_{1}^{2}(8-8t)dt$

$D=(8t-4t^2)_{0}^{1}-(8t-4t^2)_{1}^{2}$

$D=8-4+0-0-0+8-4 \ m$

$D=8 \ m$

Hence, The distance traveled by the particle is 8 m