Constant Acceleration Motion

From Constant Acceleration to Motion

When an object moves through space, its position, velocity, and acceleration are vector-valued functions of time. If its acceleration does not change, then the velocity changes at a constant rate. The position follows a quadratic formula in elapsed time.

These formulas are useful because they turn a motion problem with known initial data into a direct calculation. The initial position and velocity determine where the object is and how fast it is moving at any later time, provided its acceleration stays constant.

We will use \(\mathbf r(t)\) for position, \(\mathbf v(t)=\mathbf r'(t)\) for velocity, and \(\mathbf a(t)=\mathbf v'(t)\) for acceleration. All vectors are expressed in a fixed Cartesian coordinate system.

Let \(t_0\) be an initial time, with position \(\mathbf r_0=\mathbf r(t_0)\) and velocity \(\mathbf v_0=\mathbf v(t_0)\). Write \(\Delta t=t-t_0\) for the elapsed time. The formulas apply for times \(t\) in the interval where the acceleration remains constant.

Deriving the Velocity and Position Formulas

The velocity formula follows by accumulating the constant acceleration over the elapsed time. Then the position formula follows by accumulating the resulting velocity. The vector-valued Fundamental Theorem of Calculus applies component by component.

For the first step, the Fundamental Theorem of Calculus gives

Here \(s\) is the integration variable, and \(\mathbf a\) is fixed, so its integral over an interval of length \(t-t_0\) is \(\mathbf a(t-t_0)\). Therefore,

For position, integrate this velocity expression over the same time interval. Since \(s-t_0\) measures elapsed time at the intermediate time \(s\),

The last equality uses \(\int_{t_0}^{t}(s-t_0)\,ds=\frac12(t-t_0)^2\). Thus the position formula is

The formulas can also be checked directly. Differentiating the position expression gives \(\mathbf r'(t)=\mathbf v_0+\mathbf a(t-t_0)=\mathbf v(t)\). Differentiating the velocity expression gives \(\mathbf v'(t)=\mathbf a\). At \(t=t_0\), the elapsed time is zero, so the expressions give \(\mathbf v(t_0)=\mathbf v_0\) and \(\mathbf r(t_0)=\mathbf r_0\). This verifies both the motion and its initial data.

Reading the Formulas in Three Dimensions

Write the fixed acceleration as \(\mathbf a=(a_x,a_y,a_z)\), and write the initial position and velocity as \(\mathbf r_0=(x_0,y_0,z_0)\) and \(\mathbf v_0=(v_{0x},v_{0y},v_{0z})\). Then the position formula means

The velocity formula has the corresponding component form:

These component expressions are simply the vector formulas written coordinate by coordinate. Each coordinate uses the same elapsed time, but its own initial position, initial velocity, and acceleration component. A zero acceleration component means that velocity component remains constant; it does not require the other components to remain constant.

Worked Examples

A Common Error: Losing the One-Half

The position formula contains \(\frac12\mathbf a(\Delta t)^2\), not \(\mathbf a(\Delta t)^2\). The factor \(\frac12\) comes from integrating elapsed time: the integral of \(s-t_0\) from \(t_0\) to \(t\) is half the square of the total elapsed time.

For a concrete check, suppose \(\mathbf r_0=(0,0,0)\), \(\mathbf v_0=(1,0,0)\), and \(\mathbf a=(0,2,0)\), with \(t_0=0\). At \(t=3\), the incorrect calculation that omits the one-half gives

The correct acceleration contribution is \(\frac12(0,2,0)(9)=(0,9,0)\), so

The wrong formula also fails a derivative check: it would give a second-coordinate position term \(2t^2\), whose second derivative is \(4\), not the required acceleration component \(2\). The correct term is \(t^2\), whose second derivative is \(2\).

When the Formulas Apply

The essential condition is that the acceleration vector remains fixed over the time interval of interest. If its direction changes, even while its magnitude stays the same, the constant-vector-acceleration formulas do not apply as written. Likewise, when acceleration changes with time, one cannot substitute a single constant vector for \(\mathbf a(t)\) throughout the derivation.

The formulas also distinguish initial values from elapsed time. If the initial data are given at \(t_0\ne 0\), use \(t-t_0\), not \(t\), in both formulas. The initial position and velocity are the values at \(t_0\), so using the wrong time origin would generally fail the initial-condition check.

For dimensional consistency, if time is measured in seconds, then \(\mathbf v_0\Delta t\) has units of length, and \(\mathbf a(\Delta t)^2\) also has units of length. In the velocity equation, \(\mathbf a\Delta t\) has units of velocity. These checks are useful safeguards against mismatched quantities.