A particle is moving with constant acceleration a. Following graph shows v^2 versus x (displacement)

Q: Why did we use the third equation of motion?

Because the graph gives us a relationship between $v^2$ and $x$, and the third equation ($v^2 = u^2 + 2ax$) is the only one that directly connects these two variables without involving time.

Q: What does the y-intercept of this graph represent?

The y-intercept is the value of $v^2$ when $x = 0$. According to the equation $v^2 = u^2 + 2ax$, when $x = 0$, $v^2 = u^2$. So, the y-intercept represents the square of the initial velocity ($u^2$).

Q: Can we find the initial velocity from this graph?

Yes! The y-intercept is at $v^2 = 20$. Therefore, the initial velocity $u = \sqrt{20} = 2\sqrt{5} \text{ m/s}$.

Q: What if the graph was a curve instead of a straight line?

If the $v^2$ vs $x$ graph is a curve, it means the slope is not constant. Since the slope represents $2a$, a changing slope implies that the acceleration is not constant.

Q: Why is the slope equal to $2a$ and not just $a$?

When we differentiate $v^2 = u^2 + 2ax$ with respect to $x$, we get $\frac{d(v^2)}{dx} = 2a$. Since the slope of the graph is exactly $\frac{d(v^2)}{dx}$, it must be equal to $2a$.

Question

A particle is moving with constant acceleration $a$. Following graph shows $v^2$ versus $x$ (displacement) plot. The acceleration of the particle is ........ $	ext{m/s}^2$

Sigma Percentile · Accepted Answer

### Decoding the Graph

When you first look at the problem, you are presented with a graph plotting the square of velocity ($v^2$) against displacement ($x$). The most crucial observation here is the shape of the graph—it is a perfectly straight line. In physics, whenever you see a straight line graph, your immediate instinct should be to relate it to the standard equation of a straight line: $y = mx + c$.

Here, our y-axis represents $v^2$ and our x-axis represents $x$. So, the equation governing this motion takes the form $v^2 = mx + c$. But what do the slope ($m$) and the y-intercept ($c$) represent physically? To answer that, we need to dive into our kinematics toolkit.

### The Kinematic Connection

We need an equation that connects velocity, displacement, and constant acceleration without involving time. The third equation of motion is the perfect fit:

$$v^2 = u^2 + 2ax$$

Let's rearrange this slightly to match our straight-line format:

$$v^2 = (2a)x + u^2$$

By comparing this with $y = mx + c$, a beautiful relationship emerges. The slope of our graph, $m$, is exactly equal to twice the acceleration ($2a$). Furthermore, the y-intercept, $c$, corresponds to the square of the initial velocity ($u^2$). This means that just by finding the slope of this line, we can unlock the value of the acceleration.

### Calculating the Slope

Now, let's extract the slope from the given graph. The slope $m$ is defined as the "rise over run", or the change in the y-variable divided by the change in the x-variable:

$$m = \frac{\Delta v^2}{\Delta x}$$

We can pick any two clear points on the line to calculate this. Let's use the starting point $(0, 20)$ and the final point $(30, 80)$.

$$m = \frac{80 - 20}{30 - 0}$$

$$m = \frac{60}{30} = 2$$

### The Final Acceleration

We have established that the slope of the graph is $2$, and from our kinematic comparison, we know that the slope is equal to $2a$. Setting these equal to each other gives us a simple algebraic equation:

$$2a = 2$$

Solving for $a$, we get:

**$$a = 1 	ext{ m/s}^2$$**

And there we have it! By simply understanding the geometric properties of the graph and linking them to fundamental kinematic equations, we effortlessly arrived at the constant acceleration of the particle.

Animated Solution for Physics - Kinematics: A particle is moving with constant acceleration $a$ . Following graph shows $v^{2}$ versus $x$ (displacement) plot. The acceleration of the particle is ........ $m/s^{2}$

Enter Numerical Value:

Visualized Solution

$v^{2} vs x Graph$

$Equation of Motion$

$Comparing with Straight Line$

$Calculating Slope$

$Final Acceleration$

$Conceptual Extension$

The Sigma Insight: Motion Graphs

Decoding the Graph

The Kinematic Connection

Calculating the Slope

The Final Acceleration

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