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Subsection 2.5.1 Average Acceleration

The average rate of change of velocity during an interval can be obtained by dividing the change in velocity by the interval. Let \(v_{i,x} \) be the velocity at \(t_i\) and \(v_{f,x} \) be the velocity at \(t_f\text{,}\) then average acceleration, \(a_{\text{av},x}\text{,}\) will be

We often write this as

where \(\Delta v_x = v_{f,x} - v_{i,x}\) and \(\Delta t = t_f - t_i\text{.}\) Note that for average aceleration, we do not need to know what happened at any other instant than just the initial and final instants. For instance, for a motion shown in Figure 2.5.1, the average acceleration between A and B will be obtained from \(v_x\) at those points and time between them.

Note that \(v_x(\text{at B})\) is negative since the particle is moving towards negative \(x\) axis.

\begin{align*} \amp v_x(A) = 2.0\text{ m/s}, \\ \amp v_x(B) = - 2.0\text{ m/s}, \\ \amp a_{\text{av},x} = \frac{-2.0-2.0}{2.0} = -2.0\text{ m/s}^2. \end{align*}

Subsection 2.5.2 Instantaneous Acceleration

Often we want acceleration at a particular instant, called instantaneous acceleration, or simply, acceleration. This refers to the average acceleration between that instant, say \(t \text{,}\) and an instant very close to \(t\text{,}\) e.g., a tiny amount of time later than \(t\text{,}\) which we can indicate by \(t + \Delta t \text{.}\) Or, alternately, between \(t\) and a tiny instant before \(t\text{,}\) say at \(t-\Delta t\text{.}\)

How close instants \(t \) and \(t + \Delta t \) be? The answer is as close as possible except right at \(t\text{.}\) This is the notion of infintesimal interval already discussed in Subsection 2.4.3 on instantaneous velocity.

Denoting instantaneous acceleration for a motion along \(x \) axis by symbol \(a_x \text{,}\) we write this special averaging formally using the notion of a limit in Calculus,

which is also denoted by the derivative symbol \(dv_x/dt\text{.}\)

Graphically, this derivative can be computed by the slope of the tangent to the \(v_x \) versus \(t\) plot at the instant of interest.

Positive and Negative \(a_x\) When Moving Towards \(x = + \infty\text{:}\) When an object is moving towards \(x = + \infty\) with increasing speed, we will have \(v_x\left(\text{at } t+\Delta t\right) \gt v_x\left(\text{at } t\right)\text{,}\) which will give positive \(a_x\text{.}\) And, when the object is moving towards \(x = + \infty\) with decreasing speed, we will have \(v_x\left(\text{at } t+\Delta t\right) \lt v_x\left(\text{at } t\right)\text{,}\) which will give negative \(a_x\text{.}\)

Positive and Negative \(a_x\) When Moving Towards \(x = - \infty\text{:}\) When the object is moving towards \(x = - \infty\text{,}\) \(v_x \lt 0\text{.}\) That means that if the motion is with increasing speed, we will have \(v_x\left(\text{at } t+\Delta t\right) \lt v_x\left(\text{at } t\right)\text{,}\) which will give negative \(a_x\text{.}\) And, when the object is moving towards \(x = - \infty\) but with decreasing speed, we will have \(v_x\left(\text{at } t+\Delta t\right) \gt v_x\left(\text{at } t\right)\text{,}\) which will give positive \(a_x\text{.}\)

Subsection 2.5.3 Graphical Definition of Acceleration

Since the slope of a function is the rate of change of that function, the rate of change of velocity at a particular instant can be obtained from the slope of the tangent of the velocity versus time plot. We can take this as defining acceleration \(a_x \) for one-dimensional motion on the \(x \) axis.

If the plot of \(v_x \) versus \(t \) is linear, i.e., a straight line, then the tangent line is same as the line of the plot, hence \(a_x \) is just the slope of the line.

But, if the plot of \(v_x \) versus \(t \) is curved, then you would draw a tangent line at the instant of interest, and compute the slope of the tangent line to get \(a_x \text{.}\)

From the \(v_x \) versus \(t \) plot in Figure 2.5.5, find \(a_x \) at the following instants: (a) \(t = 1\text{ sec}\text{,}\) (b) \(t = 3\text{ sec} \text{,}\) and (c) \(t = 3.75\text{ sec} \text{.}\)

Hint

Use slopes.

Answer

(a) \(0\text{,}\) (b) \(-4\text{ m/s}^2\text{,}\) (c) \(4\text{ m/s}^2\)

Solution

(a) The \(v_x \) versus \(t\) plot through the time when \(t = 1\text{ sec}\) is a straight line. Therefore, \(a_x \) will simply be slope of this part of the plot. The slope of this line is clearly zero. Therefore, \(a_x (\text{at }t=1\text{ sec}) = 0\text{.}\)

(b) The \(v_x \) versus \(t\) plot through the time when \(t = 3\text{ sec}\) is a straight line. Therefore, \(a_x \) will simply be slope of this part of the plot. The slope of this line is obtained by picking two points on the line between \(t = 2\text{ sec}\) and \(t = 3.5\text{ sec}\text{.}\) We pick points at \(t = 2\text{ sec}\) and \(t = 3\text{ sec}\) to obtain

\begin{equation*} a_x = \dfrac{0 - 4\text{ m/s}}{3\text{ s}-2\text{ s}} = -4\text{ m/s}^2. \end{equation*}

(b) The \(v_x \) versus \(t\) plot through the time when \(t = 3.75\text{ sec}\) is a straight line. Therefore, \(a_x \) will simply be slope of this part of the plot. The slope of this line is obtained by picking two points on the line between \(t = 3.5\text{ sec}\) and \(t = 4.0\text{ sec}\text{.}\) We pick points at \(t = 3.5\text{ sec}\) and \(t = 4\text{ sec}\) to obtain

\begin{equation*} a_x = \dfrac{0 - (-2)\text{ m/s}}{4\text{ s}-3.5\text{ s}} = 4\text{ m/s}^2. \end{equation*}

Subsection 2.5.4 \(\Delta v_x\) from \(a_x\) Graphically

Since \(a_x\) is detivative of \(v_x(t)\text{,}\) a change in \(v_x\) will be given by an integral of \(a_x(t)\text{.}\) Since we can think of an integral as area under a plot of \(a_x\) between the \(a_x(t)\) and \(a_x=0\) as illustrated in Figure 2.5.6.. We have seen same relation between \(v_x\) and \(x(t)\) as we are seeing between \(a_x\) and \(v_x(t)\text{.}\)

The area under the curve is the area between the graph and \(a_x=0\) line. Therefore, from positive \(a_x\) we get positive area and for negative \(a_x\) we get negative area. Positive \(a_x\) leads to positive change in the \(v_x\) and a negative \(a_x\) to a negative change in \(v_x\text{.}\) This makes sense; say, starting \(v_x = 10\text{ m/s}\) and acceleration is constant, \(a_x=-3\text{ m/s}^2\text{.}\) Then, change in velocity in \(2\text{ sec}\) will be \(\Delta v_x = -3\times 2 = -6\text{ m/s}\text{.}\) This will give \(v_x = 4\text{ m/s}\) after \(2\text{ sec}\text{.}\)

Subsection 2.5.5 (Calculus) Analytic Definition of Acceleration

The derivative of velocity with respect to time gives the rate at which velocity changes. That is, acceleration \(a_x \) on the \(x \) axis is the derivative of the velocity \(v_x \text{.}\)

Subsection 2.5.6 (Calculus) Change in Velocity from Acceleration Analytically

The inverse problem of obtaining change in velocity, \(\Delta v_x\text{,}\) from a known acceleration, \(a_x(t) \text{,}\) is a similar problem of going from \(v_x \) to \(\Delta x\text{.}\) Therefore, the math is similar.

The inverse of Eq. (2.5.5) gives the change in velocity \(\Delta v_x = v_{f,x} - v_{i,x}\) when we integrate acceleration \(a_x \text{.}\)

In this equation, if we choose \(t_f\) to be a variable \(t\text{,}\) then we will get velocity as a function of \(t\text{.}\) Writing \(v_{f,x}\) as \(v_x(t)\) we have

where \(v_{i,x}\) is some known velocity at the initial instant and I have replaced the integration variable by symbol \(t'\) so as not to get confused by \(t\text{.}\) Integrating \(v_x\) as a function we will get the change in position.

Let us work out the case of a constant acceleration. Let us say that acceleration has a constant value \(a_x=a_0\text{,}\) initial instant \(t_i=0\text{,}\) and \(v_{i,x}=v_0\text{.}\) Let us write \(t_f-t_i\) as \(T\text{.}\) Then, we will get the following for change in \(x\text{.}\)