Velocity-Displacement

We have just seen that velocity is directly proportional to time and displacement is proportional to time squared. With a little bit of thinking, a new proportionality statement should be apparent. Change in displacement is proportional to the change in the square of velocity when acceleration is constant. This statement is particularly important for driving safety. When you double the speed of a car, it takes four times the distance to stop it. Triple the speed and you’ll need nine times the distance. Like the previous relationships, it also depends on initial velocity. Unfortunately determining the effect using reasoning alone is a real chore. Do the algebra instead.

The last two equations each described one kinematic variable as a function of time. It would be nice if we also had an equation that was independent of time. That is, we want to answer the question, “What is the relationship between velocity and displacement?” The method of doing this should be readily apparent. We’ve got to combine our first two equations of motion together in a manner that will eliminate time as a variable. The easiest way to do that is to solve one equation for time and then substitute it into the other. The second equation of motion is a quadratic and solving it for time would introduce a lot of nastiness into the algebra. It should be apparent that solving the first equation of motion [1] for time and substituting it into the second [2] will be the easier process.

Substituting Δt=(v-v₀)/a

x-x₀=v₀(v-v₀/a)+1/2a(v-v₀)²

x-x₀=(vv₀-v₀)/a+(v²-2vv₀+v²₀)/2a

2a(x-x₀)=(2vv₀-2v²₀)+(v²-2vv₀+v²₀)

As expected, displacement is proportional to velocity squared. (Initial velocity squared is just a constant that must be dealt with.) Unlike the first and second equations of motion, there is no advantage in using calculus to derive the third equation of motion. Algebra is the way to go.

Conservation of Momentum