1.3 Position and Velocity Equations

Several useful equations describing the relationship between position, velocity, and acceleration may now be introduced.

This formula is fairly straightforward. Velocity (v) is the sum of initial velocity (v₀) and the product of a constant acceleration (a) and the time elapsed while accelerating (t). If a person standing still (initial velocity = 0.0 m/s) accelerated at a rate of 0.5 m/s² for 2.0 seconds and then stopped accelerating (maintained his velocity), his velocity would be 0.0 m/s + 0.5m/s² × 2.0 s, or 1 m/s.

Perhaps this one is not so intuitive. Let’s break it apart and examine each term individually. First, position (x) is the sum of initial position (x₀) and two other terms. If you’ve already traveled from your starting point to some other point x₀, any further traveling you do will be in addition to that original displacement. The second term represents the product of initial velocity (v₀) and time elapsed (t). Take the case in which there is no acceleration. This causes the third term [(1/2)at²] to drop out. Then we are left with x = x₀ + v₀t², or position is equal to the sum of the initial position and the product of the initial velocity and the time elapsed. For example, a bicyclist traveling at 15 km/h would travel 18 km in 1.2 hours (x = 0.0 + (15 km/h)(1.2 h) = 18 km).

Now, the final case in which this formula is valid: motion involving constant acceleration. Perhaps the most comprehensive way to prove the validity of the third term is by discussing derivatives. As was mentioned earlier, velocity is the first derivative (rate of change) of position, and acceleration is the first derivative of velocity. Therefore, acceleration is the second derivative of position. To obtain the equation x = x₀ + v₀t + (1/2)at², we can integrate.

We’ll begin with an equation discussed earlier:

Now, let’s integrate back to position:

substituting the initial position for the constant, we get