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2D Motion

Position, Velocity, and Acceleration as Vectors

Kinematics Terms

Problems for Position, Velocity, and Acceleration as Vectors

The Position Function

In the last SparkNote, we discussed position functions in one dimension. The value of such a function at a particular time t 0 , x(t 0) , was an ordinary number which represented the position of the object along a single line. In two and three dimensions, however, the position of an object must be specified by a vector. We therefore need to upgrade our one- dimensional function x(t) to x(t) , so that at each moment in time the position of the object is now given in terms of a vector. Whereas x(t) was a scalar-valued function, x(t) is vector-valued. They are both, nevertheless, position functions.

As we might expect, the individual components of x(t) correspond to one-dimensional position functions in each of the two or three directions of motion. For instance, for motion in three dimensions, the components of x(t) can be labeled x(t) , y(t) , and z(t) , and correspond to one-dimensional position functions in the x -, y -, and z -directions, respectively. If we have three-dimensional motion with constant velocity, x(t) = v t , where v = (v x, v y, v z) is a constant vector, the above vector equation for x(t) breaks up into three one-dimensional equations:

x(t) = v x t, y(t) = v y t, z(t) = v z t

Note that if v y = v z = 0 , what we recover is just one-dimensional motion in the x -direction.

Position, Velocity, and Acceleration

What makes the generalization to vectors particularly simple is that the relationships between position, velocity, and acceleration stay exactly the same. Whereas before we had

v(t) = x'(t) and a(t) = v'(t) = x''(t)

now we have

v(t) = x â≤(t) and a(t) = v â≤(t) = x â≤â≤(t).

where the derivatives are taken component by component. In other words, if x(t) = (x(t), y(t), z(t)) , then x â≤(t) = (x'(t), y'(t), z'(t)) . Therefore, all the equations derived in the previous section are valid once the scalar-valued functions are turned into vector-valued ones.

As an example, consider the position function

x(t) = a t 2 + v 0 t + x 0,

where a = (0, 0, - g) , v 0 = (v x, 0, v z) , and x 0 = (0, 0, h) . The above vector equation for position can be broken down into three one-dimensional equations:

x(t) = v x t, y(t) = 0, z(t) = - gt 2 + v z t + h

The motion in the x -direction is of constant velocity, the motion in the y -direction is non-existent (so really this is a two-dimensional problem), and the motion in the z -direction looks like that of an object moving up and down near the surface of the earth (recall that g = 9.8 m/s2 is the acceleration due to gravity near the earth's surface). However, as long as the vectors a , v 0 , and x 0 are specified, all of this information can be compressed into a single vector equation.

It is important to keep in mind that, although the vector equations for kinematics look almost identical to their scalar counterparts, the range of physical phenomena that they can describe is far greater. The last example suggests that for the same object, completely different motions can be going on in the x -, y -, and z -directions, even though they are all part of one overall motion. This idea of breaking up an object's motion into components will help us analyze two- and three-dimensional motion by using ideas we've already learned from the one-dimensional case. In the next section, we put some of these methods to work when we discuss motion with constant acceleration in more than one dimension.

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