3.2 Instantaneous Velocity And Speed | University Physics Volume 1

Instantaneous Velocity

The quantity that tells us how fast an object is moving anywhere along its path is the instantaneous velocity, usually called simply velocity. It is the average velocity between two points on the path in the limit that the time (and therefore the displacement) between the two points approaches zero. To illustrate this idea mathematically, we need to express position x as a continuous function of t denoted by x(t). The expression for the average velocity between two points using this notation is $$ \overset{\text{–}}{v}=\frac{x({t}_{2})-x({t}_{1})}{{t}_{2}-{t}_{1}}$$. To find the instantaneous velocity at any position, we let $$ {t}_{1}=t $$ and $$ {t}_{2}=t+\text{Δ}t$$. After inserting these expressions into the equation for the average velocity and taking the limit as $$ \text{Δ}t\to 0$$, we find the expression for the instantaneous velocity:

$$v(t)=\underset{\text{Δ}t\to 0}{\text{lim}}\frac{x(t+\text{Δ}t)-x(t)}{\text{Δ}t}=\frac{dx(t)}{dt}.$$ Instantaneous Velocity

The instantaneous velocity of an object is the limit of the average velocity as the elapsed time approaches zero, or the derivative of x with respect to t:

$$v(t)=\frac{d}{dt}x(t).$$

Like average velocity, instantaneous velocity is a vector with dimension of length per time. The instantaneous velocity at a specific time point $$ {t}_{0} $$ is the rate of change of the position function, which is the slope of the position function $$ x(t) $$ at $$ {t}_{0}$$. (Figure) shows how the average velocity $$ \overset{\text{–}}{v}=\frac{\text{Δ}x}{\text{Δ}t} $$ between two times approaches the instantaneous velocity at $$ {t}_{0}. $$ The instantaneous velocity is shown at time $$ {t}_{0}$$, which happens to be at the maximum of the position function. The slope of the position graph is zero at this point, and thus the instantaneous velocity is zero. At other times, $$ {t}_{1},{t}_{2}$$, and so on, the instantaneous velocity is not zero because the slope of the position graph would be positive or negative. If the position function had a minimum, the slope of the position graph would also be zero, giving an instantaneous velocity of zero there as well. Thus, the zeros of the velocity function give the minimum and maximum of the position function.

Figure 3.6 In a graph of position versus time, the instantaneous velocity is the slope of the tangent line at a given point. The average velocities $$ \overset{\text{–}}{v}=\frac{\text{Δ}x}{\text{Δ}t}=\frac{{x}_{\text{f}}-{x}_{\text{i}}}{{t}_{\text{f}}-{t}_{\text{i}}} $$ between times $$ \text{Δ}t={t}_{6}-{t}_{1},\text{Δ}t={t}_{5}-{t}_{2},\text{and}\,\text{Δ}t={t}_{4}-{t}_{3} $$ are shown. When $$ \text{Δ}t\to 0$$, the average velocity approaches the instantaneous velocity at $$ t={t}_{0}$$.

Example

Finding Velocity from a Position-Versus-Time Graph

Given the position-versus-time graph of (Figure), find the velocity-versus-time graph.

Figure 3.7 The object starts out in the positive direction, stops for a short time, and then reverses direction, heading back toward the origin. Notice that the object comes to rest instantaneously, which would require an infinite force. Thus, the graph is an approximation of motion in the real world. (The concept of force is discussed in Newton’s Laws of Motion.)

Strategy

The graph contains three straight lines during three time intervals. We find the velocity during each time interval by taking the slope of the line using the grid.

Solution

Show Answer

Time interval 0 s to 0.5 s: $$ \overset{\text{–}}{v}=\frac{\text{Δ}x}{\text{Δ}t}=\frac{0.5\,\text{m}-0.0\,\text{m}}{0.5\,\text{s}-0.0\,\text{s}}=1.0\,\text{m/s}$$

Time interval 0.5 s to 1.0 s: $$ \overset{\text{–}}{v}=\frac{\text{Δ}x}{\text{Δ}t}=\frac{0.0\,\text{m}-0.0\,\text{m}}{1.0\,\text{s}-0.5\,\text{s}}=0.0\,\text{m/s}$$

Time interval 1.0 s to 2.0 s: $$ \overset{\text{–}}{v}=\frac{\text{Δ}x}{\text{Δ}t}=\frac{0.0\,\text{m}-0.5\,\text{m}}{2.0\,\text{s}-1.0\,\text{s}}=-0.5\,\text{m/s}$$

The graph of these values of velocity versus time is shown in (Figure).

Figure 3.8 The velocity is positive for the first part of the trip, zero when the object is stopped, and negative when the object reverses direction.

Significance

During the time interval between 0 s and 0.5 s, the object’s position is moving away from the origin and the position-versus-time curve has a positive slope. At any point along the curve during this time interval, we can find the instantaneous velocity by taking its slope, which is +1 m/s, as shown in (Figure). In the subsequent time interval, between 0.5 s and 1.0 s, the position doesn’t change and we see the slope is zero. From 1.0 s to 2.0 s, the object is moving back toward the origin and the slope is −0.5 m/s. The object has reversed direction and has a negative velocity.