Solving for Displacement
Solving for Displacement (Δx) and Final Position (x) From Average Velocity When Acceleration (a) is Constant
To get our first two new equations, we start with the definition of average velocity:
\(\bar{v} = \cfrac{\Delta x}{\Delta t}.\)
Substituting the simplified notation for \(\Delta x\) and \(\Delta t\) yields
\(\bar{v} = \cfrac{x \; - \; x_0}{t}.\)
Solving for \(x\) yields
\(x = x_0 + \bar{v} t,\)
where the average velocity is
\(\bar{v} = \cfrac{v_0 + v}{2} (\text{constant }a).\)
The equation \(\bar{v} = \frac{v_0 + v}{2}\) reflects the fact that, when acceleration is constant, \(v\) is just the simple average of the initial and final velocities. For example, if you steadily increase your velocity (that is, with constant acceleration) from 30 to 60 km/h, then your average velocity during this steady increase is 45 km/h. Using the equation \(\bar{v} = \frac{v_0 + v}{2}\) to check this, we see that
\(\bar{v} = \cfrac{v_0 + v}{2} = \cfrac{30\text{ km/h} + 60\text{ km/h}}{2} = 45\text{ km/h},\)
which seems logical.
Example on Calculating Displacement: How Far does the Jogger Run?
A jogger runs down a straight stretch of road with an average velocity of 4.00 m/s for 2.00 min. What is his final position, taking his initial position to be zero?
Strategy
Draw a sketch.
Image credit: OpenStax Physics
The final position \(x\) is given by the equation:
\(x = x_0 + \bar{v} t.\)
To find \(x,\) we identify the values of \(x_0, \bar{v},\) and \(t\) from the statement of the problem and substitute them into the equation.
Solution
1. Identify the knowns. \(\bar{v} = 4.00\text{ m/s},\) \(\Delta t = 2.00\text{ min},\) and \(x_0 = 0\text{ m}.\)
2. Enter the known values into the equation.
\(x = x_0 + \bar{v} t = 0 + (4.00\text{ m/s})(120\text{ s}) = 480\text{ m}\)
Discussion
Velocity and final displacement are both positive, which means they are in the same direction.
The equation \(x = x_0 + \bar{v} t\) gives insight into the relationship between displacement, average velocity, and time. It shows, for example, that displacement is a linear function of average velocity. (By linear function, we mean that displacement depends on \(\bar{v}\) rather than on \(\bar{v}\) raised to some other power, such as \(\bar{v}^2.\) When graphed, linear functions look like straight lines with a constant slope.) On a car trip, for example, we will get twice as far in a given time if we average 90 km/h than if we average 45 km/h.
There is a linear relationship between displacement and average velocity. For a given time t, an object moving twice as fast as another object will move twice as far as the other object.
This lesson is part of:
One-Dimensional Kinematics