Animated Solution for Physics - Kinematics: Relation between average velocity $v_{av}$ of a body and time $t$ is shown in the graph. If during the time interval considered, the body did not change direction of motion, draw a graph between instantaneous velocity of the body and time.

A Tale of Averages and Instants

At first glance, a graph of average velocity versus time might seem like a straightforward piece of information. However, hidden within its slopes and segments lies the dynamic, ever-changing story of instantaneous velocity. In this problem, we are tasked with extracting that hidden story. We are given a piecewise linear graph of average velocity $v_{av}$ and asked to construct the graph of instantaneous velocity $v$.

To translate average velocity into instantaneous velocity, we need a mathematical bridge. We start with the fundamental definition of average velocity:

where $s$ is the total displacement and $t$ is the total time. Rearranging this, we can express displacement as a function of time:

Now, how do we find instantaneous velocity? By definition, it is the rate of change of displacement with respect to time, $v=\frac{ds}{dt}$. Since our displacement is a product of two functions of time ($v_{av}$ and $t$), we must invoke the product rule of differentiation:

This elegant equation is our master key. It tells us that instantaneous velocity is not just the average velocity, nor is it just the slope of the average velocity graph. It is a combination of both! Let's apply this key to unlock each interval of our journey.

Interval 1: $0\le t<1$ In the first second, the average velocity graph is a straight line starting from the origin $(0,0)$ and reaching $(1,2)$. The equation of this line is simply $v_{av}=2t$. The slope, $\frac{d{v}_{av}}{dt}$, is a constant $2$. Plugging these into our master equation:

The instantaneous velocity starts at $0$ and ramps up linearly to $4\text{ m/s}$ at $t=1$.

Interval 2: $1\le t<2$ Suddenly, the average velocity levels off and becomes perfectly horizontal at $v_{av}=2$. Because it's constant, its slope is zero ($\frac{d{v}_{av}}{dt}=0$). Our master equation simplifies beautifully:

Notice the dramatic shift! At exactly $t=1$, the instantaneous velocity plummets from $4\text{ m/s}$ down to $2\text{ m/s}$. This creates a discontinuous jump in our graph.

Interval 3: $2\le t<3$ The journey takes another turn. The average velocity graph climbs again, connecting the points $(2,2)$ and $(3,3)$. The slope of this segment is $\frac{3-2}{3-2}=1$. Using the point-slope form, the equation is $v_{av}=t$. Applying our master equation once more:

At $t=2$, the velocity abruptly jumps back up from $2\text{ m/s}$ to $4\text{ m/s}$, and then climbs steadily to $6\text{ m/s}$ at $t=3$.

Interval 4: $3\le t\le 4$ In the final stretch, the average velocity plateaus again at $v_{av}=3$. Just like in the second interval, the slope is zero.

At $t=3$, we witness our final discontinuous drop, as the velocity falls from $6\text{ m/s}$ to a steady $3\text{ m/s}$.

When we piece these segments together, we get a fascinating, disconnected graph. These sudden vertical jumps in instantaneous velocity imply that the object underwent infinite acceleration for an infinitesimally small moment—a physical impossibility in the real world, but a brilliant mathematical idealization that deeply tests our understanding of calculus in kinematics. By trusting the math over intuition, we successfully unveiled the hidden velocity!