Lectures on Physics has been derived from Benjamin Crowell's Light and Matter series of free introductory textbooks on physics. See the editorial for more information....

Algebraic Results for Constant Acceleration

r / The shaded area tells us how far an object moves while accelerating at a constant rate.

Although the area-under-the-curve technique can be applied to any graph, no matter how complicated, it may be laborious to carry out, and if fractions of rectangles must be estimated the result will only be approximate. In the special case of motion with constant acceleration, it is possible to find a convenient shortcut which produces exact results. When the acceleration is constant, the v - t graph is a straight line, as shown in the figure. The area under the curve can be divided into a triangle plus a rectangle, both of whose areas can be calculated exactly: A = bh for a rectangle and A = bh/2 for a triangle. The height of the rectangle is the initial velocity, vo, and the height of the triangle is the change in velocity from beginning to end, Δv. The object's Δx is therefore given by the equation Δx = voΔt + ΔvΔt/2. This can be simplified a little by using the definition of acceleration, a = Δv/Δt, to eliminate Δv, giving

Δx = voΔt + 1/2 aΔt2 . [motion with constant acceleration]

Since this is a second-order polynomial in Δt, the graph of Δx versus Δt is a parabola, and the same is true of a graph of x versus t - the two graphs differ only by shifting along the two axes. Although I have derived the equation using a figure that shows a positive vo, positive a, and so on, it still turns out to be true regardless of what plus and minus signs are involved.

Another useful equation can be derived if one wants to relate the change in velocity to the distance traveled. This is useful, for instance, for finding the distance needed by a car to come to a stop. For simplicity, we start by deriving the equation for the special case of vo = 0, in which the final velocity vf is a synonym for Δv. Since velocity and distance are the variables of interest, not time, we take the equation Δx = 1/2 aΔt2 and use Δt = Δv/a to eliminate Δt. This gives Δx = (Δv)2/a, which can be rewritten as

vf2 = 2aΔx . [motion with constant acceleration, vo = 0]

For the more general case where , we skip the tedious algebra leading to the more general equation,

vf2 = vo2 + 2aΔx . [motion with constant acceleration]

To help get this all organized in your head, first let's categorize the variables as follows:

Variables that change during motion with constant acceleration:

x ,v, t

Variable that doesn't change:


If you know one of the changing variables and want to find another, there is always an equation that relates those two:

The symmetry among the three variables is imperfect only because the equation relating x and t includes the initial velocity.

There are two main difficulties encountered by students in applying these equations:

  • The equations apply only to motion with constant acceleration. You can't apply them if the acceleration is changing.

  • Students are often unsure of which equation to use, or may cause themselves unnecessary work by taking the longer path around the triangle in the chart above. Organize your thoughts by listing the variables you are given, the ones you want to find, and the ones you aren't given and don't care about.

Saving an old lady.

→ Solved problem: A stupid celebration page 125, problem 15 → Solved problem: Dropping a rock on Mars page 125, problem 16 → Solved problem: The Dodge Viper page 126, problem 18 → Solved problem: Half-way sped up page 126, problem 22

Discussion Questions

A Check that the units make sense in the three equations derived in this section.
B In chapter 1, I gave examples of correct and incorrect reasoning about proportionality, using questions about the scaling of area and volume. Try to translate the incorrect modes of reasoning shown there into mistakes about the following question: If the acceleration of gravity on Mars is 1/3 that on Earth, how many times longer does it take for a rock to drop the same distance on Mars?

Last Update: 2010-11-11