Instantaneous Velocity And Speed

The average velocity tells us how fast an object has been moving over a given time interval but does not tell us how fast it moves at different instants of time during that interval. For this, we define instantaneous velocity or simply velocity v at an instant t. 

The velocity at an instant is defined as the limit of the average velocity as the time interval ∆t becomes infinitesimally small. In other words,

$v=\lim_{\Delta t\rightarrow 0}\frac{\Delta x}{\Delta t}$                        (1a)

$v=\frac{dx}{dt}$                                                                                   (1b)

where the symbol $\lim_{\Delta t\rightarrow 0}$ stands for the operation of taking limit as ∆t→0 of the quantity on its right. In the language of calculus, the quantity on the right hand side of Eq. (1a) is the differential coefficient of x with respect to t and is denoted by $\frac{dx}{dt}$ (see Appendix 3.1). It is the rate of change of position with respect to time, at that instant. 

We can use Eq. (1a) for obtaining the value of velocity at an instant either graphically or numerically. Suppose that we want to obtain graphically the value of velocity at time t = 4 s (point P) for the motion of the car represented in Fig. 1. 

Fig 1: Position-time graph of a car.

The figure has been redrawn in Fig. 2 choosing different scales to facilitate the calculation. Let us take ∆t = 2 s centred at t = 4 s. Then, by the definition of the average velocity, the slope of line $P_1P_2$ (Fig. 2) gives the value of average velocity over the interval 3 s to 5 s. Now, we decrease the value of ∆t from 2 s to 1 s. Then line $P_1P_2$ becomes $Q_1Q_2$ and its slope gives the value of the average velocity over the interval 3.5 s to 4.5 s. In the limit ∆t → 0, the line $P_1P_2$ becomes tangent to the position-time curve at the point P and the velocity at t = 4 s is given by the slope of the tangent at that point. It is difficult to show this process graphically. But if we use numerical method to obtain the value of the velocity, the meaning of the limiting process becomes clear. 

Fig 2: Determining velocity from position-time graph. Velocity at t = 4 s is the slope of the tangent to the graph at that instant.

For the graph shown in Fig. 2, x = 0.08 $t^3$. Table 3.1 gives the value of ∆x/∆t calculated for ∆t equal to 2.0 s, 1.0 s, 0.5 s, 0.1 s and 0.01 s centred at t = 4.0 s. The second and third columns give the value of  $t_1=\left(t-\frac{\Delta t}{2}\right)$ and $t_2=\left(t+\frac{\Delta t}{2}\right)$  and the fourth and the fifth columns give the corresponding values of x, i.e. x ($t_1$) = 0.08 $t_1^3$  and x ($t_2$) = 0.08 $t_2^3$. The sixth column lists the difference ∆x = x($t_2$) – x($t_1$) and the last column gives the ratio of ∆x and ∆t, i.e. the average velocity corresponding to the value of ∆t listed in the first column. 

We see from Table 1 that as we decrease the value of ∆t from 2.0 s to 0.010 s, the value of the average velocity approaches the limiting value 3.84 m/s which is the value of velocity at t = 4.0 s, i.e. the value of $\frac{dx}{dt}$  at t = 4.0 s. In this manner, we can calculate velocity at each instant for motion of the car shown in Fig. 3.3. For this case, the variation of velocity with time is found to be as shown in Fig. 3.

Fig 3: Velocity–time graph corresponding to motion shown in Fig. 3.3.

The graphical method for the determination of the instantaneous velocity is always not a convenient method. For this, we must carefully plot the position–time graph and calculate the value of average velocity as ∆t becomes smaller and smaller. It is easier to calculate the value of velocity at different instants if we have data of positions at different instants or exact expression for the position as a function of time. Then, we calculate ∆x/∆t from the data for decreasing the value of ∆t and find the limiting value as we have done in Table 3.1 or use differential calculus for the given expression and calculate $\frac{dx}{dt}$ at different instants as done in the following example.

Table 1: Limiting value of ∆x/∆t at t = 4 s

Example 1

The position of an object moving along x-axis is given by x = a + b$t^2$ where a = 8.5 m, b = 2.5 ms$^{–2}$ and t is measured in seconds. What is its velocity at t = 0 s and t = 2.0 s. What is the average velocity between t = 2.0 s and t = 4.0 s ?

Answer 

In notation of differential calculus, the velocity is

$v=\frac{dx}{dt}=\frac{d}{dt}\left(a+bt^2\right)=2b=5t$ m/s

At t = 0 s, v = 0 m/s and at t = 2.0 s, v = 10 m/s.

average velocity $= \frac{x(4.0)-x(2.0)}{4.0 - 2.0}$

$= \frac{a+16b-a-4b}{2.0}=6.0 \times b$

= 6.0 x 2.5 = 15 m/s

From Fig. 3, we note that during the period t =10 s to 18 s the velocity is constant. Between period t =18 s to t = 20 s, it is uniformly decreasing and during the period t = 0 s to t = 10 s, it is increasing. 

Note that for uniform motion, velocity is the same as the average velocity at all instants. 

Instantaneous speed or simply speed is the magnitude of velocity. For example, a velocity of + 24.0 m/s and a velocity of – 24.0 m/s — both have an associated speed of 24.0 m/s. It should be noted that though average speed over a finite interval of time is greater or equal to the magnitude of the average velocity, instantaneous speed at an instant is equal to the magnitude of the instantaneous velocity at that instant. Why so ?