The figure shows the position-time (x-t) graph of one-dimensional motion of a body of mass 0.4 kg. T

Q: Why is the velocity constant in each segment?

Because the $x-t$ graph is made of straight lines. A straight line has a constant slope, and since slope equals velocity, the velocity must be constant.

Q: Why do we take the magnitude at the end?

The question specifically asks for the 'magnitude of each impulse'. The negative sign only indicates the direction of the impulse relative to the chosen coordinate system.

Q: What does the negative sign in the impulse mean?

It means the impulse is directed opposite to the initial direction of motion. It's the force required to stop the object and send it back the way it came.

Q: Is the impulse the same at $t=4$ s?

Yes, the magnitude is exactly the same. At $t=4$ s, the velocity goes from $-1 \text{ m/s}$ to $+1 \text{ m/s}$, resulting in an impulse of $+0.8 \text{ N-s}$. The magnitude remains $0.8 \text{ N-s}$.

Q: Can we use $F = ma$ to solve this?

No, because the force is applied for an infinitesimally small time (at the exact corner of the graph), making the acceleration technically infinite. In such collision-like scenarios, we must use the concept of impulse.

Question

The figure shows the position-time (x-t) graph of one-dimensional motion of a body of mass 0.4 kg. The magnitude of each impulse is

Options:

(A) 0.4 N-s

(B) 0.8 N-s

(C) 1.6 N-s

(D) 0.2 N-s

Sigma Percentile · Accepted Answer

### Analyzing the Setup

When we look at a position-time ($x-t$) graph, the most crucial piece of information it holds is its slope. The slope of an $x-t$ graph represents the velocity of the object. In this specific problem, the graph is composed entirely of straight-line segments. A straight line has a constant slope, which immediately tells us that the object is moving with a constant velocity during each of these intervals.

### The Logic of the Kink

Notice the sharp corners, or "kinks," at the peaks and troughs of the graph (for example, at $t = 2$ s, $t = 4$ s, etc.). At these exact moments, the slope changes abruptly from positive to negative, or vice versa.

In the physical world, an abrupt, instantaneous change in velocity means that an infinite acceleration occurred over an infinitesimally small time interval. This is the hallmark of an **impulse**. Impulse ($J$) is defined as the change in momentum of an object:

$$J = \Delta p = m(v_f - v_i)$$

### Calculating the Velocities

To find the impulse at the first peak ($t = 2$ s), we need to determine the velocity just before and just after this moment.

For the first segment (from $t = 0$ to $t = 2$ s), the object moves from $x = 0$ to $x = 2$ m. The initial velocity $v_1$ is the slope of this line:

$$v_1 = \frac{\Delta x}{\Delta t} = \frac{2 - 0}{2 - 0} = 1 	ext{ m/s}$$

For the second segment (from $t = 2$ to $t = 4$ s), the object moves from $x = 2$ m back to $x = 0$ m. The final velocity $v_2$ is the slope of this descending line:

$$v_2 = \frac{\Delta x}{\Delta t} = \frac{0 - 2}{4 - 2} = -1 	ext{ m/s}$$

### The Master Equation: Impulse-Momentum Theorem

Now that we have our velocities, we can apply the impulse-momentum theorem. The mass of the body is given as $m = 0.4 	ext{ kg}$. Substituting our values into the impulse equation:

$$J = m(v_2 - v_1)$$

$$J = 0.4(-1 - 1)$$

$$J = 0.4(-2) = -0.8 	ext{ N-s}$$

### Final Calculation

The negative sign simply indicates that the direction of the impulse is opposite to the initial direction of motion (it forced the object to turn around). However, the question specifically asks for the **magnitude** of the impulse.

Taking the absolute value, we get:

$$|J| = 0.8 	ext{ N-s}$$

This elegant problem beautifully connects the geometric properties of a graph to the dynamic physical reality of forces and momentum.

Animated Solution for Physics - Laws of Motion: The figure shows the position-time (x-t) graph of one-dimensional motion of a body of mass 0.4 kg. The magnitude of each impulse is

Select Answer:

Visualized Solution

$x - t Graph Analysis$

$The Physics of Kinks$

$v_{1} Calculation$

$v_{2} Calculation$

$Applying Impulse Theorem$

$Final Answer$

$The Way Forward$

The Sigma Insight: Inertia, Momentum, and Impulse

Analyzing the Setup

The Logic of the Kink

Calculating the Velocities

The Master Equation

Final Calculation

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