Solving for Final Position

Solving for Final Position When Velocity is Not Constant (a ≠ 0)

We can combine the equations above to find a third equation that allows us to calculate the final position of an object experiencing constant acceleration. We start with

\(v = v_0 + at.\)

Adding \(v_0\) to each side of this equation and dividing by 2 gives

\(\cfrac{v_0 + v}{2} = v_0 + \cfrac{1}{2}at.\)

Since \(\cfrac{v_0 + v}{2} = \bar{v}\) for constant acceleration, then

\(\bar{v} = v_0 + \cfrac{1}{2}at.\)

Now we substitute this expression for \(\bar{v}\) into the equation for displacement, \(x = x_0 + \bar{v} t,\) yielding

\(x = x_0 + v_0t + \cfrac{1}{2}at^2 (\text{constant }a).\)

Example on Calculating Displacement of an Accelerating Object: Dragsters

Dragsters can achieve average accelerations of 26.0 m/s². Suppose such a dragster accelerates from rest at this rate for 5.56 s. How far does it travel in this time?

U.S. Army Top Fuel pilot Tony “The Sarge” Schumacher begins a race with a controlled burnout. Drag racing is a type of motor racing in which automobiles or motorcycles (usually specially prepared for the purpose) compete, usually two at a time, to be first to cross a set finish line. Top Fuel dragsters are the quickest accelerating racing vehicles in the world and the fastest sanctioned category of drag racers, with the fastest competitors reaching speeds of 335 miles per hour (539 km/h) and finishing the 1,000 foot (305 m) runs in 3.7 seconds. Image credit: Lt. Col. William Thurmond. Photo Courtesy of U.S. Army.

Strategy

Draw a sketch.

Image credit: OpenStax Physics

We are asked to find displacement, which is \(x\) if we take \(x_0\) to be zero. (Think about it like the starting line of a race. It can be anywhere, but we call it 0 and measure all other positions relative to it.) We can use the equation \(x = x_0 + v_0t + \frac{1}{2}at^2\) once we identify \(v_0, a,\) and \(t\) from the statement of the problem.

Solution

1. Identify the knowns. Starting from rest means that \(v_0 = 0,\) \(a\) is given as \(26.0\mathrm{ m/s^2}\) and \(t\) is given as 5.56 s.

2. Plug the known values into the equation to solve for the unknown \(x\):

\(x = x_0 + v_0t + \cfrac{1}{2}at^2.\)

Since the initial position and velocity are both zero, this simplifies to

\(x = \cfrac{1}{2}at^2.\)

Substituting the identified values of \(a\) and \(t\) gives

\(x = \cfrac{1}{2} \left ( 26.0\mathrm{\; m/s^2} \right ) (5.56\text{ s})^2,\)

yielding

\(x = 402\text{ m}.\)

Discussion

If we convert 402 m to miles, we find that the distance covered is very close to one quarter of a mile, the standard distance for drag racing. So the answer is reasonable. This is an impressive displacement in only 5.56 s, but top-notch dragsters can do a quarter mile in even less time than this.

What else can we learn by examining the equation \(x = x_0 + v_0t + \frac{1}{2}at^2?\) We see that:

displacement depends on the square of the elapsed time when acceleration is not zero. In the example above, the dragster covers only one fourth of the total distance in the first half of the elapsed time
if acceleration is zero, then the initial velocity equals average velocity \((v_0 = \bar{v})\) and \(x = x_0 + v_0t + \frac{1}{2}at^2\) becomes \(x = x_0 + v_0t\)

Solving for Final Position