Position x(t) for Particle Moving at Acceleration a(t)=6t-2

[the_d]

The acceleration of a particle moving along
the x-axis at time t is given by a(t) = 6t - 2.
If the velocity is 25 when t = 3 and the po-
sition is 10 when t = 1, then finnd the position
x(t).

 

[dextercioby]

Do you know calculus??Apply the definition of both acceleration wrt velocity & velocity wrt coordinate.
Take care of the conditions which will fix integration constants...

It shouldn't be 2 difficult.

Daniel.

 

[xanthym]

The acceleration of a particle moving along
the x-axis at time t is given by a(t) = 6t - 2.
If the velocity is 25 when t = 3 and the po-
sition is 10 when t = 1, then finnd the position
x(t).

Let a(t) be the acceleration defined above, v(t) the velocity, and x(t) the x-axis position. By definition:

$$v(t) = \int a(t) dt = \int (6t - 2) dt = 3t^2 - 2t + C_1$$
$$x(t) = \int v(t) dt = \int (3t^2 - 2t + C_1) dt = t^3 - t^2 + C_1*t + C_2$$

where C1 and C2 are constants. We determine the latter constants by using values given in the problem statement:

$$v(3) = 25 = 3*(3)^2 -2*(3) + C_1$$
$$C_1 = 4$$

$$x(1) = 10 = (1)^3 - (1)^2 + (4)*(1) + C_2$$
$$C_2 = 6$$

$$\color{red} x(t) = t^3 - t^2 + 4t + 6$$