The velocity of an object, in general, changes
during its course of motion. How to describe this
change? Should it be described as the rate of
change in velocity with distance or with time ?
This was a problem even in Galileo’s time. It was
first thought that this change could be described
by the rate of change of velocity with distance.
But, through his studies of motion of freely falling
objects and motion of objects on an inclined
plane, Galileo concluded that the rate of change
of velocity with time is a constant of motion for
all objects in free fall. On the other hand, the
change in velocity with distance is not constant
– it decreases with the increasing distance of fall.
This led to the concept of acceleration as the rate
of change of velocity with time.
The average acceleration a over a time
interval is defined as the change of velocity
divided by the time interval:
$\bar{a}=\frac{v_2-v_1}{t_2-t_1}=\frac{\Delta v}{\Delta t}$ (1)
where $v_2$ and $v_1$ are the instantaneous velocities
or simply velocities at time $t_2$ and $t_1$ . It is the
average change of velocity per unit time. The SI
unit of acceleration is ms$^{–2}$.
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Fig.1
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On a plot of velocity versus time, the average acceleration is the slope of the straight line connecting the points corresponding to ($v_2$, $t_2$) and ($v_1$, $t_1$). The average acceleration for velocity-time graph shown in Fig. 1 for different time intervals 0 s - 10 s, 10 s – 18 s, and 18 s – 20 s are:0 s - 10 s → $\bar{a}=\frac{(24-0) \ m.s^{-1}}{(10-0) \ s}$ = 2.4 $m.s^{-1}$
10 s - 18 s → $\bar{a}=\frac{(24-24) \ m.s^{-1}}{(18-10) \ s}$ = 0 $m.s^{-1}$
18 – 20 s → $\bar{a}=\frac{(0-24) \ m.s^{-1}}{(20-18) \ s}$ = -12 $m.s^{-1}$
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| Fig. 2: Acceleration as a function of time for motion
represented in Fig. 1. |
Instantaneous acceleration is defined in the same
way as the instantaneous velocity:
$a=\lim_{\Delta t\rightarrow 0}\frac{\Delta v}{\Delta t}=\frac{\Delta v}{\Delta t}$ (1)
The acceleration at an instant is the slope of
the tangent to the v–t curve at that instant. For
the v–t curve shown in Fig. 1, we can obtain
acceleration at every instant of time. The
resulting a – t curve is shown in Fig. 2. We see that the acceleration is nonuniform over the
period 0 s to 10 s. It is zero between 10 s and
18 s and is constant with value –12 m$s^{–2}$ between 18 s and 20 s. When the acceleration
is uniform, obviously, it equals the average
acceleration over that period.
Since velocity is a quantity having both
magnitude and direction, a change in velocity
may involve either or both of these factors.
Acceleration, therefore, may result from a
change in speed (magnitude), a change in
direction or changes in both. Like velocity,
acceleration can also be positive, negative or
zero. Position-time graphs for motion with
positive, negative and zero acceleration are
shown in Figs. 3(a), (b) and (c), respectively.
Note that the graph curves upward for positive
acceleration; downward for negative
acceleration and it is a straight line for zero
acceleration. As an exercise, identify in Figure,
the regions of the curve that correspond to these
three cases.
Although acceleration can vary with time,
our study in this chapter will be restricted to
motion with constant acceleration. In this case,
the average acceleration equals the constant
value of acceleration during the interval. If the
velocity of an object is vo
at t = 0 and v at time t,
we have
$\bar{a}=\frac{v-v_0}{t-0}$ or $v = v_0+at$ (2)
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| Fig. 3: Position-time graph for motion with
(a) positive acceleration; (b) negative
acceleration, and (c) zero acceleration. |
Let us see how velocity-time graph looks like
for some simple cases. Fig. 4 shows velocity-time graph for motion with constant acceleration
for the following cases:
(a) An object is moving in a positive direction
with a positive acceleration, for example
the motion of the car in Figure between
t = 0 s and t = 10 s.
(b) An object is moving in positive direction
with a negative acceleration, for example,
motion of the car in Figure between
t = 18 s and 20 s.
(c) An object is moving in negative direction
with a negative acceleration, for example
the motion of a car moving from O in Figure in negative x-direction with
increasing speed.
(d) An object is moving in positive direction
till time $t_1$, and then turns back with the
same negative acceleration, for example
the motion of a car from point O to point
Q in Figure till time $t_1$ with decreasing
speed and turning back and moving with
the same negative acceleration.
An interesting feature of a velocity-time graph
for any moving object is that the area under the
curve represents the displacement over a
given time interval. A general proof of this statement requires use of calculus. We can,
however, see that it is true for the simple case of
an object moving with constant velocity u. Its
velocity-time graph is as shown in Fig. 5.
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| Fig.4 |
Fig 4: Velocity–time graph for motions with
constant acceleration. (a) Motion in positive
direction with positive acceleration,
(b) Motion in positive direction with
negative acceleration, (c) Motion in negative
direction with negative acceleration,
(d) Motion of an object with negative
acceleration that changes direction at time $t_1$. Between times 0 to $t_1$, its moves in
positive x - direction and between $t_1$ and $t_2$ it moves in the opposite direction.
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| Fig 5: Area under v–t curve equals displacement
of the object over a given time interval. |
The v-t curve is a straight line parallel to the
time axis and the area under it between t = 0
and t = T
is the area of the rectangle of height u
and base T. Therefore, area = u × T = uT which
is the displacement in this time interval. How
come in this case an area is equal to a distance?
Think! Note the dimensions of quantities on
the two coordinate axes, and you will arrive at
the answer.
Note that the x-t, v-t, and a-t graphs shown
in several figures in this chapter have sharp
kinks at some points implying that the
functions are not differentiable at these
points. In any realistic situation, the
functions will be differentiable at all points
and the graphs will be smooth.
What this means physically is that
acceleration and velocity cannot change
values abruptly at an instant. Changes are
always continuous.
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