Question

A particle moving along the x-axis has its velocity described by the function vx =2t2m/s, where t is in s. its initial position is x0 = 2.8 m at t0 = 0 s . at 1.3 s , what is the particle's position?

Answer 1

At time t = 1.3 s, the position of the particle is 4.3 m.

Velocity of a particle is the rate of change of its displacement.

Therefore, the velocity of the particle $v_x$ along the x direction is given by,

$v_x = \frac{dx}{dt}$

The change in the position of the particle is given by,

$dx = vdt$

Integrate the equation.

$x=\int {v_x} \, dt$

Substitute $v_x =2t^2$ and simplify.

$x=\int {v_x} \, dt =\int {2t^2} \, dt = \frac{2}{3} t^3+C$

Calculate the constant of integration C by applying initial conditions, when  t =0, x= 2.8 m.

$x=\frac{2}{3} t^3 + 2.8$

Substitute 1.3 s for t and calculate the value of x.

$x=\frac{2}{3} t^3 + 2.8\\ x_1_._3 =\frac{2}{3}(1.3)^3+2.8= 4.2647 m\\ =4.3 m (2 sf)$

The position of the particle after 1.3 s is 4.3 m