Move an object that starts at the origin, with an initial velocity of < 1,0,0> m/s with a net force of < 0,-1,3> n acting on it. allow the motion to continue for a time interval of 1.00 s. the object must show the path it has taken. the object will have a mass of 0.5 kg.

position vector  R = (Vox t, -t2, 3t2) m

position at 1.00s  R = (1, -1, 3) m

Explanation:

In this case we must use Newton's second law to find acceleration and kinematic equations to find the position of the body. Note that the system is in three dimensions (x, y, z) so we will work each axis independently. Let's start by writing the values they give us on the axes explicitly.

      V = 1 i ^ + 0 j ^ + 0 k ^

      F = 0 i ^ -1 j ^ +3 K ^

Let's start by analyzing the movement in the X-axis. Since the force is zero, the speed remains constant, the trajectory is

      X = Vox t

      X = 1 t

We continue with the Y axis. Let's look for acceleration with Newton's second law

      F = ma

      a = F / m

      a = -1 /0.5

      a = - 2m / s²

With the kinematic equation

      Y = Voy t + ½ a t²

      Y = - ½ 2 t²

      Y = -1 t²

Let's calculate on the Z axis

      a = F / m

      a = 3 / 0.5

      a = 6 m /s²

      Z = Voz t + ½ to t²

      Z = ½ 6 t²

      Z = 3 t²2

Let's build the position vector for the object

      R =  t i ^ - t2 j ^ + 3 t2 k ^ [m]

We can also write them the way

      R = ( t, -t2, 3t2) m

We can see that the movement is uniform (a straight line) on the X axis and is accelerated (abolished torque) on the Y and Z axes,

Finally, let's calculate the position for the time given to us by 1.00 s

          R = 1 i ^ - 12 j ^ + 3 12 k ^ [m]

          R = i ^ - j ^ +3 k ^

          R = (1, -1, 3) m