3.4 Motion with constant acceleration  (Page 11/10)

You might guess that the greater the acceleration of, say, a car moving away from a stop sign, the greater the car’s displacement in a given time. But, we have not developed a specific equation that relates acceleration and displacement. In this section, we look at some convenient equations for kinematic relationships, starting from the definitions of displacement, velocity, and acceleration. We first investigate a single object in motion, called single-body motion. Then we investigate the motion of two objects, called two-body pursuit problems .

Notation

First, let us make some simplifications in notation. Taking the initial time to be zero, as if time is measured with a stopwatch, is a great simplification. Since elapsed time is $\text{Δ}t=t_{\text{f}}-t_0$ , taking $t_0=0$ means that $\text{Δ}t=t_{\text{f}}$ , the final time on the stopwatch. When initial time is taken to be zero, we use the subscript 0 to denote initial values of position and velocity. That is, $x_0$ is the initial position and $v_0$ is the initial velocity . We put no subscripts on the final values. That is, t is the final time , x is the final position , and v is the final velocity . This gives a simpler expression for elapsed time, $\text{Δ}t=t$ . It also simplifies the expression for x displacement, which is now $\text{Δ}x=x-x_0$ . Also, it simplifies the expression for change in velocity, which is now $\text{Δ}v=v-v_0$ . To summarize, using the simplified notation, with the initial time taken to be zero,

where the subscript 0 denotes an initial value and the absence of a subscript denotes a final value in whatever motion is under consideration.

We now make the important assumption that acceleration is constant . This assumption allows us to avoid using calculus to find instantaneous acceleration. Since acceleration is constant, the average and instantaneous accelerations are equal—that is,

Thus, we can use the symbol a for acceleration at all times. Assuming acceleration to be constant does not seriously limit the situations we can study nor does it degrade the accuracy of our treatment. For one thing, acceleration is constant in a great number of situations. Furthermore, in many other situations we can describe motion accurately by assuming a constant acceleration equal to the average acceleration for that motion. Lastly, for motion during which acceleration changes drastically, such as a car accelerating to top speed and then braking to a stop, motion can be considered in separate parts, each of which has its own constant acceleration.

Displacement and position from velocity

To get our first two equations, we start with the definition of average velocity:

Substituting the simplified notation for $\text{Δ}x$ and $\text{Δ}t$ yields

Solving for x gives us

where the average velocity is

The equation $\stackrel{\text{–}}{v}=\frac{v_0+v}{2}$ reflects the fact that when acceleration is constant, v is just the simple average of the initial and final velocities. [link] illustrates this concept graphically. In part (a) of the figure, acceleration is constant, with velocity increasing at a constant rate. The average velocity during the 1-h interval from 40 km/h to 80 km/h is 60 km/h: