Motion Equations for Constant Acceleration in One Dimension

Read this text for more examples and practice on how to solve motion equations for constant acceleration.

Solving for Final Velocity

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 \mathrm{~km} / \mathrm{h} than if we average 45 \mathrm{~km} / \mathrm{h}.

Line graph showing displacement in meters versus average velocity in meters per second. The line is straight with a positive slope. Displacement x increases linearly with increase in average velocity v.

Figure 2.27 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.

Solving for Final Velocity

We can derive another useful equation by manipulating the definition of acceleration.

a=\frac{\Delta v}{\Delta t}.

Substituting the simplified notation for \Delta v and \Delta t gives us

a=\frac{v-v_{0}}{t}(\text { constant } a).

Solving for v yields

v=v_{0}+a t(\operatorname{constant} a).

Example 2.9 Calculating Final Velocity: An Airplane Slowing Down after Landing

An airplane lands with an initial velocity of 70.0 \mathrm{~m} / \mathrm{s} and then decelerates at 1.50 \mathrm{~m} / \mathrm{s}^{2} for 40.0 \mathrm{~s}. What is its final velocity?

Strategy

Draw a sketch. We draw the acceleration vector in the direction opposite the velocity vector because the plane is decelerating.

Velocity vector arrow pointing toward the right in the positive x direction. Initial velocity equals seventy meters per second. Final velocity equals question mark. An acceleration vector arrow pointing toward the left labeled a equals negative 1 point 50 meters per second squared.

Figure 2.28

Solution

1. Identify the knowns. v_{0}=70.0 \mathrm{~m} / \mathrm{s}, a=-1.50 \mathrm{~m} / \mathrm{s}^{2}, t=40.0 \mathrm{~s}.

2. Identify the unknown. In this case, it is final velocity, v_{\mathrm{f}}.

3. Determine which equation to use. We can calculate the final velocity using the equation v=v_{0}+a t.

4. Plug in the known values and solve.

v=v_{0}+a t=70.0 \mathrm{~m} / \mathrm{s}+\left(-1.50 \mathrm{~m} / \mathrm{s}^{2}\right)(40.0 \mathrm{~s})=10.0 \mathrm{~m} / \mathrm{s}

Discussion

The final velocity is much less than the initial velocity, as desired when slowing down, but still positive. With jet engines, reverse thrust could be maintained long enough to stop the plane and start moving it backward. That would be indicated by a negative final velocity, which is not the case here.

An airplane moving toward the right at two points in time. At time equals 0 the velocity vector arrow points toward the right and is labeled seventy meters per second. The acceleration vector arrow points toward the left and is labeled negative 1 point 5 meters per second squared. At time equals forty seconds, the velocity arrow is shorter, points toward the right, and is labeled ten meters per second. The acceleration vector arrow is still pointing toward the left and is labeled a equals negative 1 point 5 meters per second squared.

Figure 2.29 The airplane lands with an initial velocity of 70.0 m/s and slows to a final velocity of 10.0 m/s before heading for the terminal. Note that the acceleration is negative because its direction is opposite to its velocity, which is positive.

In addition to being useful in problem solving, the equation v=v_{0}+a t gives us insight into the relationships among velocity, acceleration, and time. From it we can see, for example, that

  • final velocity depends on how large the acceleration is and how long it lasts
  • if the acceleration is zero, then the final velocity equals the initial velocity \left(v=v_{0}\right), as expected (i.e., velocity is constant )
  • if a is negative, then the final velocity is less than the initial velocity

(All of these observations fit our intuition, and it is always useful to examine basic equations in light of our intuition and experiences to check that they do indeed describe nature accurately).

Making Connections: Real-World Connection

Space shuttle blasting off at night.

Figure 2.30 The Space Shuttle Endeavor blasts off from the Kennedy Space Center in February 2010.

An intercontinental ballistic missile (ICBM) has a larger average acceleration than the Space Shuttle and achieves a greater velocity in the first minute or two of flight (actual ICBM burn times are classified - short-burn-time missiles are more difficult for an enemy to destroy). But the Space Shuttle obtains a greater final velocity, so that it can orbit the earth rather than come directly back down as an ICBM does. The Space Shuttle does this by accelerating for a longer time.

Solving for Final Position When Velocity is Not Constant (a \neq 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

\frac{v_{0}+v}{2}=v_{0}+\frac{1}{2} at.

Since \frac{v_{0}+v}{2}=\bar{v} for constant acceleration, then

\bar{v}=v_{0}+\frac{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_{0} t+\frac{1}{2} a t^{2}(\operatorname{constant} a).

Example 2.10 Calculating Displacement of an Accelerating Object: Dragsters

Dragsters can achieve average accelerations of 26.0 \mathrm{~m} / \mathrm{s}^{2}. Suppose such a dragster accelerates from rest at this rate for 5.56 s. How far does it travel in this time?

Dragster accelerating down a race track.

Figure 2.31 U.S. Army Top Fuel pilot Tony "The Sarge" Schumacher begins a race with a controlled burnout.

Strategy

Draw a sketch.

Acceleration vector arrow pointing toward the right in the positive x direction, labeled a equals twenty-six point 0 meters per second squared. x position graph with initial position at the left end of the graph. The right end of the graph is labeled x equals question mark.

Figure 2.32

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_{0} t+\frac{1}{2} a t^{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} / \mathrm{s}^{2} and t is given as 5.56 \mathrm{~s}.

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

x=x_{0}+v_{0} t+\frac{1}{2} a t^{2}.

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

x=\frac{1}{2} a t^{2}.

Substituting the identified values of a and t gives

x=\frac{1}{2}\left(26.0 \mathrm{~m} / \mathrm{s}^{2}\right)(5.56 \mathrm{~s})^{2}.

yielding

x=402 \mathrm{~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_{0} t+\frac{1}{2} a t^{2} ? We see that:

  • displacement depends on the square of the elapsed time when acceleration is not zero. In Example 2.10, 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 \left(v_{0}=\bar{v}\right) and x=x_{0}+v_{0} t+\frac{1}{2} a t^{2} becomes x=x_{0}+v_{0} t

Solving for Final Velocity When Velocity is Not Constant (a \neq 0)

A fourth useful equation can be obtained from another algebraic manipulation of previous equations. If we solve v=v_{0}+a t for t, we get

t=\frac{v-v_{0}}{a}.

Substituting this and \bar{v}=\frac{v_{0}+v}{2} into x=x_{0}+\bar{v} t, we get

v^{2}=v_{0}^{2}+2 a\left(x-x_{0}\right)(\operatorname{constant} a).

Example 2.11 Calculating Final Velocity: Dragsters

Calculate the final velocity of the dragster in Example 2.10 without using information about time.

Strategy

Draw a sketch.

Acceleration vector arrow pointing toward the right, labeled twenty-six point zero meters per second squared. Initial velocity equals 0. Final velocity equals question mark.

Figure 2.33


The equation v^{2}=v_{0}^{2}+2 a\left(x-x_{0}\right) is ideally suited to this task because it relates velocities, acceleration, and displacement, and no time information is required.

Solution

1. Identify the known values. We know that v_{0}=0, since the dragster starts from rest. Then we note that x-x_{0}=402 \mathrm{~m} (this was the answer in Example 2.10). Finally, the average acceleration was given to be a=26.0 \mathrm{~m} / \mathrm{s}^{2}

2. Plug the knowns into the equation v^{2}=v_{0}^{2}+2 a\left(x-x_{0}\right) and solve for v.

v^{2}=0+2\left(26.0 \mathrm{~m} / \mathrm{s}^{2}\right)(402 \mathrm{~m}).

Thus

v^{2}=2.09 \times 10^{4} \mathrm{~m}^{2} / \mathrm{s}^{2}.

To get v, we take the square root:

v=\sqrt{2.09 \times 10^{4} \mathrm{~m}^{2} / \mathrm{s}^{2}}=145 \mathrm{~m} / \mathrm{s}.

Discussion

145 m/s is about 522 km/h or about 324 mi/h, but even this breakneck speed is short of the record for the quarter mile. Also, note that a square root has two values; we took the positive value to indicate a velocity in the same direction as the acceleration.

An examination of the equation v^{2}=v_{0}^{2}+2 a\left(x-x_{0}\right) can produce further insights into the general relationships among physical quantities:

  • The final velocity depends on how large the acceleration is and the distance over which it acts
  • For a fixed deceleration, a car that is going twice as fast doesn't simply stop in twice the distance-it takes much further to stop. (This is why we have reduced speed zones near schools).