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 ?
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