A particle is initially at the position r0 = 3i + 2j (m) and its acceleration is a = -10j (m/s2). The particle has an initial velocity given by: v0 = 2i + 2j (m/s). Find the velocity, position and acceleration as a function of time. Express the trajectory of the particle in the form y(x).
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First we are going to draw the Cartesian axes that we will use to solve the problem as well as the initial conditions of it.
The acceleration vector of the particle (which is not drawn to scale) points in the negative direction of the vertical axis.
Notice that we have moved the axes to the end of the position vector r0, in order to draw the initial velocity vector v0 (in magenta).
The acceleration of the particle is an given in this problem and does not depend on time.
To calculate the velocity vector we integrate the definition of the acceleration:
If you want to see in more detail how the limits of the integral are chosen, check out Problem 1.
Integrating both members of the equation:
We can also express separately each of the Cartesian components of the velocity vector:
Both ways of expressing the velocity vector are equivalent; in the first one, the vector is expressed as a function of the unit vectors. The second one gives the Cartesian components. The first one is the most used in Physics.
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To calculate the position vector we start from the definition of the velocity vector and integrate it using the velocity vector that we have just calculated:
And substituting the initial position vector:
To express the trajectory of the particle in the form y(x) we first express the position vector in parametric form:
Now we isolate the variable (t) in the first equation and substitute into the second:
Therefore the particle describes a parabolic motion. Below you have a plot of the parabola.
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