JEE PYQ: Motion in a Straight Line - Question ID d5c326eec609 (JEE Main 2023)

In kinematics, the relationships between displacement ($s$), velocity ($v$), and acceleration ($a$) are derived from calculus. The key formulas are:

• Velocity is the time derivative of displacement: $v = \frac{ds}{dt}$ Conversely, displacement is the integral of velocity over time: $s = \int v \, dt$ Geometrically, this integral corresponds to the area under the velocity-time graph.
• Acceleration is the time derivative of velocity: $a = \frac{dv}{dt}$ Therefore, the change in velocity is the integral of acceleration over time: $\Delta v = v(t_2) - v(t_1) = \int_{t_1}^{t_2} a \, dt$ This integral is the area under the acceleration-time graph.

However, a crucial distinction must be made:

• The area under the velocity-time graph gives the displacement, not necessarily the distance. Distance is the total path length and is always non-negative, whereas displacement can be negative if the body reverses direction.
• The area under the acceleration-time graph directly gives the change in velocity, which is always correct.

Analyzing Statement I:

Statement I claims: “Area under velocity-time graph gives the distance travelled by the body in a given time.”

1. From calculus, the integral of velocity over time is displacement: $s = \int_{t_1}^{t_2} v \, dt$
2. If the velocity is always positive (or always negative), then the magnitude of displacement equals the distance travelled. However, if the velocity changes sign (i.e., the body reverses direction), the integral gives the net displacement, which is less than the total distance.
3. Example: A body moves forward with $v = +5\,$m/s for 2 s, then backward with $v = -3\,$m/s for 2 s.
   • Displacement = $(5 \times 2) + (-3 \times 2) = 10 - 6 = 4\,$m.
   • Distance = $5 \times 2 + 3 \times 2 = 10 + 6 = 16\,$m.
   The area under the $v$-$t$ graph is 4 m, but the distance is 16 m.
4. Hence, Statement I is incorrect.

Analyzing Statement II:

Statement II claims: “Area under acceleration-time graph is equal to the change in velocity in the given time.”

1. From the definition of acceleration: $a = \frac{dv}{dt} \implies dv = a \, dt$
2. Integrating both sides from $t_1$ to $t_2$: $\int_{v(t_1)}^{v(t_2)} dv = \int_{t_1}^{t_2} a \, dt$ $v(t_2) - v(t_1) = \int_{t_1}^{t_2} a \, dt$
3. The right-hand side is precisely the area under the $a$-$t$ graph between $t_1$ and $t_2$.
4. Therefore, Statement II is true.

Conclusion:

Statement I is incorrect, but Statement II is true. The correct option is C.

Students often confuse displacement with distance. The area under the $v$-$t$ graph gives displacement, not distance, unless the velocity never changes sign. Always check whether the motion involves a reversal in direction. For the $a$-$t$ graph, the area always gives the change in velocity, so no such ambiguity exists.

Exam Tip: When in doubt, sketch a simple $v$-$t$ graph with a sign change to verify the distinction between displacement and distance.