From Position to Acceleration
A parametrized particle has a position that changes as time passes. If its coordinates at time \(t\) are \(x(t)\) and \(y(t)\), then the position is represented by \(\mathbf{r}(t)=\langle x(t),y(t)\rangle\). Differentiating each coordinate gives the horizontal and vertical components of velocity. Differentiating those components again gives the components of acceleration.
Acceleration therefore describes how velocity changes, component by component. A positive horizontal acceleration means the horizontal velocity component is increasing at that instant; it does not necessarily mean that the particle is moving to the right. The same distinction applies to vertical motion.
Here \(x'(t)\) and \(y'(t)\) are the velocity components, while \(x''(t)\) and \(y''(t)\) are the acceleration components. If position is measured in meters and time in seconds, velocity has units of meters per second and acceleration has units of meters per second squared.
How to Interpret Each Component
The horizontal acceleration \(x''(t)\) measures the instantaneous rate of change of horizontal velocity \(x'(t)\). The vertical acceleration \(y''(t)\) measures the instantaneous rate of change of vertical velocity \(y'(t)\). The sign of an acceleration component tells whether its corresponding velocity component is increasing or decreasing at that instant.
| Acceleration component | Meaning for the corresponding velocity component |
|---|---|
| \(x''(t)>0\) | The horizontal velocity \(x'(t)\) is increasing at that instant. |
| \(x''(t)<0\) | The horizontal velocity \(x'(t)\) is decreasing at that instant. |
| \(x''(t)=0\) | The horizontal velocity has zero instantaneous rate of change at that instant. |
| \(y''(t)>0\) | The vertical velocity \(y'(t)\) is increasing at that instant. |
| \(y''(t)<0\) | The vertical velocity \(y'(t)\) is decreasing at that instant. |
| \(y''(t)=0\) | The vertical velocity has zero instantaneous rate of change at that instant. |
“Increasing” and “decreasing” describe the numerical values of the velocity components, not whether the particle’s coordinates are increasing. For example, a horizontal velocity of \(5\) meters per second can be decreasing while still being positive. Acceleration equal to zero at one instant also does not, by itself, show that velocity is constant over an interval.
Worked Example: Finding Acceleration at a Given Time
Worked Example: Finding Acceleration at a Given Time
A particle’s position, in meters, is \(x(t)=t^3-2t^2+t\) and \(y(t)=2t^2-3t\), where \(t\) is measured in seconds. Find its velocity and acceleration components at \(t=2\), and interpret the acceleration.
Differentiate each position coordinate to obtain the velocity components:
At \(t=2\), the velocity components are
Differentiate the velocity components to find acceleration:
Thus, at \(t=2\),
The velocity at that instant is \(\langle 5,5\rangle\) meters per second, and the acceleration is \(\langle 8,4\rangle\) meters per second squared. Both acceleration components are positive, so both velocity components are increasing at \(t=2\).
Check: Differentiating the original coordinates once gives \(x'(t)=3t^2-4t+1\) and \(y'(t)=4t-3\); differentiating those expressions gives \(x''(t)=6t-4\) and \(y''(t)=4\), as used above.
Why Acceleration Is the Second Derivative of Position
The definition follows directly from the meaning of velocity and acceleration. Velocity is the rate of change of position with respect to time. Acceleration is the rate of change of velocity with respect to time. Applying that definition separately to the two coordinates gives the component formulas.
Differentiate the velocity vector component by component. Provided the coordinate functions have second derivatives, this gives
This also explains why the units change at each step. Differentiating position in meters with respect to time in seconds gives meters per second; differentiating again gives meters per second per second, or meters per second squared.
Acceleration and Velocity Changes over an Interval
Acceleration describes an instantaneous rate, but it can also be used to find a velocity change over a time interval. If \(x''\) is continuous on \([t_1,t_2]\), the Fundamental Theorem of Calculus gives the change in horizontal velocity as the integral of horizontal acceleration. The corresponding statement holds for the vertical components.
The formulas follow by applying the Fundamental Theorem of Calculus to each velocity component: an antiderivative of \(x''(t)\) is \(x'(t)\), and an antiderivative of \(y''(t)\) is \(y'(t)\). Evaluating each antiderivative at the interval endpoints gives the stated differences. If acceleration is positive throughout an interval, the corresponding velocity component increases over that interval; if it is negative throughout, that velocity component decreases.
Worked Example: Finding Velocity Changes from Acceleration
Worked Example: Finding Velocity Changes from Acceleration
A particle has position \(x(t)=t^3-3t\) and \(y(t)=-t^2+4t\), with time in seconds and position in meters. Find its acceleration components, then determine the change in each velocity component from \(t=0\) to \(t=2\).
First differentiate the position coordinates:
Differentiate again:
Over \(0\leq t\leq 2\), the horizontal acceleration \(6t\) is nonnegative, and the vertical acceleration is constantly negative. The velocity changes can be calculated by integrating acceleration:
So the horizontal velocity component changes by \(12\) meters per second, while the vertical velocity component changes by \(-4\) meters per second.
Check: From the velocity formulas, \(x'(0)=-3\) and \(x'(2)=3(2)^2-3=9\), so \(x'(2)-x'(0)=9-(-3)=12\). Also, \(y'(0)=4\) and \(y'(2)=-4+4=0\), so \(y'(2)-y'(0)=0-4=-4\). These agree with the integrals.
Worked Example: Acceleration with Trigonometric Coordinates
Worked Example: Acceleration with Trigonometric Coordinates
A particle’s position in meters is \(x(t)=4\cos t\), \(y(t)=3\sin t\), with \(t\) in seconds. Find the velocity and acceleration components at \(t=\frac{\pi}{3}\), then interpret the signs of the acceleration components.
Differentiate each coordinate to find velocity:
At \(t=\frac{\pi}{3}\), use \(\sin\frac{\pi}{3}=\frac{\sqrt{3}}{2}\) and \(\cos\frac{\pi}{3}=\frac{1}{2}\):
Differentiate velocity to find acceleration:
Substitute \(t=\frac{\pi}{3}\):
Both acceleration components are negative. Therefore the horizontal velocity component and the vertical velocity component are each decreasing at this instant. The velocity components themselves have different signs: the horizontal component is negative and the vertical component is positive. The acceleration signs describe how those values are changing, not the direction of either component.
Check: Differentiating \(-4\sin t\) gives \(-4\cos t\), and differentiating \(3\cos t\) gives \(-3\sin t\). Substitution of \(t=\frac{\pi}{3}\) produces the acceleration values above.
A Common Mistake: Confusing Position and Velocity
A frequent error is to interpret positive acceleration as meaning that the position coordinate is increasing. Acceleration concerns the rate of change of velocity, not the rate of change of position. For example, in the polynomial example above, at \(t=0\) the horizontal velocity is \(x'(0)=-3\), while the horizontal acceleration is \(x''(0)=0\). At \(t=1\), the horizontal velocity is \(x'(1)=3-3=0\), and the horizontal acceleration is \(x''(1)=6\). The positive acceleration at \(t=1\) says the horizontal velocity is increasing at that instant; it does not mean the horizontal velocity has a positive value at that exact instant.
Another error is to use a derivative with respect to the wrong variable. For motion parametrized by time, acceleration components come from differentiating velocity with respect to \(t\). A second derivative of one coordinate with respect to the other coordinate, such as \(\frac{d^2y}{dx^2}\), describes a feature of a curve and is not the particle’s vertical acceleration.
Check Your Understanding
Use componentwise differentiation and interpretation to answer these questions.
- If \(x(t)=2t^3-t\) and \(y(t)=t^2+5t\), find the acceleration components at \(t=1\).
- If \(x''(t)<0\) on an interval, what does this imply about the horizontal velocity component on that interval?
- Can a velocity component be positive while its acceleration component is negative? Explain what the signs mean.
- If \(y''(t)=0\) at one instant, what can and cannot be concluded about vertical velocity?
- If the vertical acceleration is constantly \(3\) meters per second squared for \(2\) seconds, what is the change in vertical velocity over that interval?
