Instantaneous Velocity and Speed


A. Instantaneous Velocity and Speed

1. Instantaneous Velocity and Speed

You have now seen two ways to describe how fast something moves: average velocity and average speed, both of which are measured over a time interval  \( \Delta t  \). However, the phrase “how fast” more commonly refers to how fast a particle is moving at a given instant – its instantaneous velocity (or simply velocity) v.

The velocity at any instant is obtained from the average velocity by shrinking the time interval  \( \Delta t  \) closer and closer to 0. As  \( \Delta t  \) dwindles, the average velocity approaches a limiting value, which is the velocity at that instant:  \( v=\underset{\Delta t\to 0}{\mathop{lim }}\,\frac{\Delta x}{\Delta t}=\frac{dx}{dt} \)   (2-4)

Note that v is the rate at which position x is changing with time at a given instant; that is, v is the derivative of x with repect to t. Also note that v at any instant is the slope of the position – time curve at the point representing that instant. Velocity is another vector quantity and thus has an associated direction.

Speed is the magnitude of velocity; that is, speed is velocity that has been stripped of any indication of direction, either in words or via an algebraic sign. (Caution: Speed and average speed can be quite different.) A velocity of +5 m/s and one of -5 m/s both have an associated speed of 5 m/s. The speedometer in a car measures speed, not velocity (it cannot determine the direction).

2. Sample problem – Velocity and slope of versus t, elevator cab

Figure 2-6a is an x(t) plot for an elevator cab that is initially stationary, then moves upward (which we take to be the postitive direction of x), and then stops. Plot v(t).

Key idea

We can find the velocity at any time from the slope of the x(t) curve at that time.

Calculations:

The slope of x(t), and so also the velocity, is zero in the intervals from 0 to 1 s and from 9 s on, so then the cab is stationary. During the interval bc, the slope is constant and nonzero, so then the cab moves with contant velocity. We calculate the slope of x(t) then as

 \( \frac{\Delta x}{\Delta t}=v=\frac{24\text{ }m-4.0\text{ }m}{8.0\text{ }s-3.0\text{ }s}=+4.0\text{ }m/s  \)

Figure 2-6 (a) The x(t) curve for an elevator cab that moves upward along an x axis. (b) The v(t) curve for the cab. Note that it is the derivative of the x(t) curve ( \( v=\frac{dx}{dt} \)). (c) The a(t) curve for the cab. It it the derivative of the v(t) curve ( \( a=\frac{dv}{dt} \)). The stick figures along them bottom suggest how a passenger’s body might fell during the accelerations.

The plus sign indicates that the cab is moving in the positive x direction. These intervals (where v = 0 and v = 4 m/s) are plotted in Fig.2-6b. In addition, as the cab initially begins to move and then later slows to a stop, v varies as indicated in the intervals 1 s to 3 s and 8 s to 9 s. Thus, Fig.2-6b is the required plot. (Figure 2-6c is considered in Module 2-3.)

Given a v(t) graph such as Fig.2-6b, we could “work backward” to produce the shape of the associated x(t) graph (Fig.2-6a). However, we would not know the actual values for x at various times, because the v(t) graph indicates only changes in x. To find such a change in x during any interval, we must, in the language of calculus, calculate the area “under the curve” on the v(t) graph for that interval.

For example, during the interval 3 s to 8 s in which the cab has a velocity of 4.0 m/s, the change in x is

 \( \Delta x=(4.0\text{ }m/s)(8.0\text{ }s-3.0\text{ }s)=+20\text{ }m  \).

(This area is positive because the v(t) curve is above the t axis.) Figure 2-6a shows that x does indeed increase by 20 m in that interval. However, Fig.2-6b does not tell us the values of x at the beginning and end of the interval. For that, we need additional information, such as the value of x at some instant.

B. Problems and Solutions

Example 1. An electron moving along the x axis has a position given by \( x=16t{{e}^{-t}}\text{ }m \), where t is in seconds. How far is the electron from the orgin when it momentarily stops?

Solution:

Using the general property  \( \frac{d}{dx}\exp (bx)=b\exp (bx) \), we write

\(v=\frac{dx}{dt}=\left( \frac{d(19t)}{dt} \right)\cdot {{e}^{-t}}+(19t)\cdot \left( \frac{d{{e}^{-t}}}{dt} \right)\)

If a concern develops about the appearance of an argument of the exponential (-t) apparently having units, then an explicit factor of  \( \frac{1}{T} \) where T = 1 second can be inserted and carried through the computation (which does not change our answer).

The result of this differentiation is

 \( v=16\left( 1-t \right){{e}^{-t}} \)

Example 2. With t and v in SI units (s and m/s, respectively). We see that this function is zero when t = 1 s. Now that we know when it stops, we find out where it stops by plugging our result t = 1 into the given function  \( x=16t{{e}^{-t}} \) with x in meters. Therefore, we find x = 5.9 m.

a) If a particle’s position is given by \( x=4-12t+3{{t}^{2}} \) (where t is in seconds and x is in meters), what is its velocity at t = 1 s?

b) Is it moving in the positive or negative direction of x just then?

c) What is its speed just then?

d) Is the speed increasing or decreasing just then? (Try answering the next two questions without further calculation.)

e) Is there ever an instant when the velocity is zero? If so, give the time t; if not, answer no.

f) Is there a time after t = 3 s when the particle is moving in the negative direction of x? If so, give the time t; if not, answer no.

Solution:

We use Eq.2-4 to solve the problem.

a) The velocity of the particle is

\(v=\frac{dx}{dt}=\frac{d}{dt}\left( 4-12t+3{{t}^{3}} \right)=-12+6t\)

Thus, at t = 1 s, the velocity is  \( v=\left( -12+(6)(1) \right)=-6\text{ }m/s  \).

b) Since v < 0, it is moving in the \( -x \) direction at t = 1 s.

c) At t = 1 s, the speed is \( \left| v \right|=6\text{ }m/s \).

d) For \( 0<t<2\text{ }s, \left| v \right| \) decreases until it vanishes. For \( 2<t<3\text{ }s  \),  \( \left| v \right| \) increases from zero to the value it had in part (c). Then,  \( \left| v \right| \) is larger than that value for  \( t>3\text{ }s  \).

e) Yes, since v smoothly changes from negative values (consider the t = 1 result) to positive (note that as \(t\to +\infty \), we have \(v\to +\infty \)). One can check that v = 0 when t = 2 s.

f) In fact, from \( v=-12+6t \), we know that v > 0 for t > 2 s.

Example 3. The position function x(t) of a particle moving along an x axis is \( x=4.0-6.0{{t}^{2}} \), with x in meters and t in seconds. (a) At what time and (b) where does the particle (momentarily) stop? At what (c) negative time and (d) positive time does the particle pass through the origin? (e) Graph x versus t for the range  \( -5\text{ }s  \) to  \( +5\text{ }s  \). (f) To shift the curve rightward on the graph, should we include the term +20t or the term -20t in x(t)? (g) Does that inclusion increase or decrease the value of x at which the particle momentarily stops?

Solution:

We use the functional notation x(t), v(t), and a(t) in this solution, where the latter two quantities are obtained by differentiation:

 \( v(t)=\frac{dx(t)}{dt}=-12t  \) and  \( a(t)=\frac{dv(t)}{dt}=-12 \)

With SI units understood.

a) From v(t) = 0 we find it is (momentarily) at rest at t = 0.

b) We obtain x(0) = 4.0 m.

c) and (d) Requiring x(t) = 0 in the expression \(x(t)=4.0-6.0{{t}^{2}}\) leads to \(t=\pm 0.82\text{ }s\) for the times when the particle can be found passing through the origin.

e) We show both the asked-for graph (on the left) as well as the “shifted” graph that is relevant to part (f). In both cases, the time axis is given by \( -3\le t\le 3 \) (SI units understood).

f) We arrived at the graph on the right (shown above) by adding 20t to the x(t) expression.

g) Examining where the slopes of the graphs become zero, it is clear that the shift causes the v = 0 point to correspond to a larger value of x (the top of the second curve shown in part (e) is higher than that of the first).

Example 4. The position of a particle moving along the x axis is given in centimeters by \( x=9.75+1.50{{t}^{3}} \), where t is in seconds. Calculate (a) the average velocity during the time interval t = 2.00 s to t = 3.00 s; (b) the instantaneous velocity at t = 2.00s; (c) the instantaneous velocity at t = 3.00 s; (d) the instantaneous velocity at t = 2.50 s; and (e) the instantaneous velocity when the particle is midway between its positions at t = 2.00 s and t = 3.00 s. (f) Graph x versus t and indicate your answers graphically.

Solution:

We use Eq.2-2 for average velocity and Eq.2-4 for instantaneous velocity, and work with distances in centimeters and times in seconds.

a) We plug into the given equation for x for t = 2.00 s and t = 3.00 s and obtain \( {{x}_{2}}=21.75\text{ }cm \) and  \( {{x}_{3}}=50.25\text{ }cm  \), respectively. The average velocity during the time interval \( 2.00\le t\le 3.00\text{ }s \) is

 \( {{v}_{avg}}=\frac{\Delta x}{\Delta t}=\frac{50.25\text{ }cm-21.75\text{ }cm}{3.00\text{ }s-2.00\text{ }s} \)

Which yields  \( {{v}_{avg}}=28.5\text{ }cm/s  \).

b) The instantaneous velocity is \( v=\frac{dx}{dt}=4.5{{t}^{2}} \), which, at time t = 2.00 s, yields \( v=(4.5){{(2.00)}^{2}}=18.0\text{ }cm/s  \)

c) At t = 3.00 s, the instantaneous velocity is \( v=(4.5){{(3.00)}^{2}}=40.5\text{ }cm/s \)

d) At t = 2.50 s, the instantaneous velocity is \( v=(4.5){{(2.50)}^{2}}=28.1\text{ }cm/s \).

e) Let tm stand for the moment when the particle is midway between x2 and x3 (that is, when the particle is at \( {{x}_{m}}=\frac{{{x}_{2}}+{{x}_{3}}}{2}=36\text{ }cm \)).

Therefore,  \( {{x}_{m}}=9,75+1.5t_{m}^{3}\Rightarrow {{t}_{m}}=2.596 \)

in seconds. Thus, the instantaneous speed at this time is \( v=4.5{{(2.596)}^{2}}=30.3\text{ }cm/s \).

f) The answer to part (a) is given by the slope of the straight line between t = 2 and t = 3 in this x-vs-t plot. The answers to parts (b), (c), (d), and (e) correspond to the slopes of tangent lines (not shown but easily imagined) to the curve at the appropriate points.

 

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