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8.2: A Special Case: Constant Acceleration

Many physical situations can be usefully modeled by assuming the acceleration is constant. In this model, there are simple algebraic relations among the motion variables that make it very easy to get quantitative answers. Even when \(a\) is not constant over the entire range of motion, it is often possible to model the motion by considering \(a\) to be constant over select intervals of time.

The algebraic equations for constant acceleration can be obtained by starting with \(a\)= constant and integrating to get \(v\) and then integrating again to get \(x\). These relations can be checked by differentiating the resulting expression for \(x\) twice and seeing if you get back the constant value for a. Remember, these equations are valid only for situations in which a is constant.

\[x(t) = x_0 + v_{x 0}t + \frac{1}{2} a_{x 0}t^2, \label{Eq1}\]

\[vx(t) = v _{x 0} + a _{x 0} t, \]

\[ax(t) = a _{x 0} ,  \]

where \(x_0\) and \(v _{x 0}\) are constants and represent the position and velocity at the time \(t = 0\) and \(a _{x 0}\) is the constant value of the acceleration for all times.

Sometimes when we are talking about motion in a vertical direction, we might use the letter y instead of x. In that case, Equation \(\ref{Eq1}\) becomes,

\[y(t ) = y_0 + v _{y 0} t + \frac{1}{2} a_{y 0}t^2 . \]

Of course, which letters we use makes no real difference as long as we think about what the symbols actually mean in each particular situation.

Remember! These equations apply only for situations which can be modeled with a constant acceleration. At other times (as when we write Newton’s 2nd law: \(\sum F = ma\) ), the acceleration is not assumed to be constant, unless it turns out that the net force is constant. One common situation when the acceleration is constant occurs when the net force is simply equal to the gravitational force of the Earth on an object near the surface of the Earth. In this case, the net force is approximately constant, so the acceleration is constant. However, there are many situations where acceleration is definitely not constant. A mass hanging on a spring oscillating up and down represents a common situation where the acceleration is certainly not constant; in this situation, \(\sum F\) oscillates from up to down as the mass moves from down to up.

A Common Constant-Acceleration Example

As a first application of Newton’s 2nd law, we examine a common example of constant acceleration: the interesting case of an object falling near the surface of the Earth. If the viscous force of the air on the object is sufficiently small compared to the pull of the Earth, we can model the situation as if only one force acts on the object: the attraction of the Earth on the object–the gravity force.

This way of modeling the forces applies whether the object is simply dropped or is thrown down or even up! Once it has left our hand, the only force acting on it is the result of the only object it is interacting with, namely the Earth. Thus, the only force is the attraction of the Earth on the object. As long as the object is not touching another object, there are no electrical or magnetic forces acting on the object, and air friction is sufficiently small so that we can neglect it, the only force acting on the object is the pull of the Earth. The object is said to be in “free fall.” When you jump up in the air a few inches off the floor, you are in free fall, because the gravitational force of the Earth on you is effectively the only force acting on you!

The force diagram (Figure 8.2.1) for an object in free fall is very simple: the single force by the Earth acting down on the object. The gravity force of the Earth on an object is simply equal to the product of the mass of the object and the proportionality constant relating the mass of an object to the gravitational force at that location, \(g\): \(F_{Earth ~on~ object} = mg\)

Figure 8.2.1: Force diagram of object in free-fall

So, applying Newton’s 2nd law to this object in free fall, we have

\[\sum F_y=m a_y\]

\(where~ \sum F_y =~ the~ gravity~ force~ of ~the ~Earth~ on~ the~ object.\)

Now the pull of the Earth points down (toward the center of the Earth). We are taking the positive direction as up right now, so the gravitational force of the Earth on the object is \(-mg\). Then substituting \(-mg\) for \(\sum F_y\) in Newton’s 2nd law, we get

\[-m g = m a_y.\]

Solving for the acceleration, a, we get

\[a_{free-fall} = -g.\]

For any object in free fall on the Earth, regardless of what its mass is, its acceleration will be equal to the quantity we have been calling g, and will point down towards the center of the Earth. The object can, of course, have any velocity—up, down, sideways, or something in between. The restriction on the object’s motion imposed by the 2nd law says nothing regarding the objects velocity, only its acceleration, and that acceleration will always be equal to the gravitational field strength. Near the surface of the Earth, the gravitational field strength is fairly constant and points toward the center of the Earth. Sometimes, \(g\) is referred to as the “acceleration due to gravity.” It is more correct to say that the acceleration of an object in free fall has the magnitude \(g\). The actual value of \(g\) depends, of course, on the properties of the particular planet: its mass and density, and on how far the object is from the center of the planet. (Newton also worked this out—his universal law of gravity, discussed in the Chapter 6.)

Quick Digression on Units

We just saw that the quantity \(g\), which we called the gravitational field strength and which has units of N/kg, is also an acceleration, which in SI is measured in m/s2. Apparently, in SI units, a Newton per kilogram is the same as a meter per second squared. Let's start from the beginning.

The fundamental units in SI in mechanics are:

  • mass: kg
  • length: m
  • time: s
All other units in mechanics are combinations of these basic three.
Quantity Unit Basic Units Relating Law
Force \(N\) \(kg \times m/s^2\) \(\sum F=ma\)
Energy \(J~or~N\times m\) \(kg \times m^2/s^2\) \(W=\int F_{\parallel}dx\)
Velocity \(m/s\) \(m/s\) \(v=dx/dt\)
Acceleration \(m/s^2\) \(m/s^2\) \(a=d^2x/dt^2\)

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