JEE PYQ: Motion in a Straight Line - Question ID a19fe5c73565 (JEE Main 2021)

Core Formula & Concept:

In kinematics, the relationship between velocity and acceleration is fundamental. Acceleration is defined as the rate of change of velocity with respect to time. Mathematically, this is expressed as: $a = \frac{dv}{dt}$ where: - $a$ is the acceleration, - $v$ is the velocity, - $t$ is the time. The velocity-time ($v$-$t$) graph visually represents how velocity changes over time. The slope of the tangent to the $v$-$t$ graph at any point gives the instantaneous acceleration at that time. Thus:

• If the $v$-$t$ graph is a straight line, the acceleration is constant (slope is constant).
• If the $v$-$t$ graph is curved, the acceleration varies (slope changes).
• A positive slope indicates positive acceleration, a negative slope indicates negative acceleration (deceleration), and a zero slope indicates zero acceleration.

Step-by-Step Derivation:

Let's analyze the given velocity-time graph (shape AMB) and derive the corresponding acceleration-time graph step-by-step.

Step 1: Analyze the Velocity-Time Graph

The given $v$-$t$ graph has the following characteristics:

• From point A to M: The graph is a straight line with a positive slope. This means the velocity is increasing at a constant rate, implying constant positive acceleration.
• At point M: The slope of the $v$-$t$ graph changes abruptly. The graph transitions from a straight line with a positive slope to a straight line with a negative slope.
• From point M to B: The graph is a straight line with a negative slope. This means the velocity is decreasing at a constant rate, implying constant negative acceleration.

Step 2: Determine the Acceleration from the Slope

Since acceleration is the slope of the $v$-$t$ graph:

• From A to M: The slope is positive and constant. Thus, the acceleration is a positive constant.
• From M to B: The slope is negative and constant. Thus, the acceleration is a negative constant.

Step 3: Sketch the Acceleration-Time Graph

Based on the above observations:

• For the time interval from A to M, the acceleration-time ($a$-$t$) graph will be a horizontal line above the time axis (since acceleration is positive and constant).
• At point M, the acceleration abruptly changes from a positive constant to a negative constant. This is represented by a vertical drop in the $a$-$t$ graph.
• For the time interval from M to B, the $a$-$t$ graph will be a horizontal line below the time axis (since acceleration is negative and constant).

Step 4: Match with Given Options

Now, let's compare this derived $a$-$t$ graph with the provided options:

• Option A: Shows a continuous curve, which is incorrect because the acceleration is piecewise constant, not continuously varying.
• Option B: Shows a single horizontal line, which is incorrect because the acceleration changes sign.
• Option C: Shows a horizontal line above the time axis, followed by a vertical drop, and then a horizontal line below the time axis. This matches our derived $a$-$t$ graph perfectly.
• Option D: Shows a horizontal line with a sudden jump upward, which is incorrect because the acceleration changes from positive to negative, not the other way around.

Thus, the correct option is C.

Common Traps & Exam Tip:

Students often make the following mistakes in such questions:

1. Misinterpreting the Slope: Some students confuse the slope of the $v$-$t$ graph with the value of velocity itself. Remember, acceleration is the slope, not the height of the $v$-$t$ graph.
2. Ignoring Discontinuities: The abrupt change in slope at point M implies a discontinuity in the $a$-$t$ graph. Students might incorrectly draw a smooth curve instead of a vertical drop.
3. Sign Errors: Failing to recognize that a negative slope in the $v$-$t$ graph corresponds to negative acceleration. This can lead to selecting an option where the acceleration does not change sign.
4. Overcomplicating the Graph: Some students assume the $a$-$t$ graph must be a curve, even when the $v$-$t$ graph consists of straight lines. Always remember that straight lines in the $v$-$t$ graph imply constant acceleration.

Exam Tip: When solving such problems, always:

• Break the $v$-$t$ graph into segments based on changes in slope.
• Determine the acceleration for each segment by calculating the slope.
• Sketch the $a$-$t$ graph segment by segment, ensuring to represent discontinuities correctly.
• Compare your sketch with the given options to identify the correct one.