JEE PYQ: Motion in a Straight Line - Question ID 29b985827956 (JEE Main 2017)

Core Formula & Concept:

When a body is thrown vertically upwards, it moves under the influence of Earth's gravitational force, which acts downward. This results in a constant acceleration directed opposite to the initial velocity. The key concepts and formulas involved are:

• Velocity-Time Relationship: For motion with constant acceleration, the velocity $v$ at any time $t$ is given by: $v = u + at$ where:
  • $u$ = initial velocity (positive, since the body is thrown upwards)
  • $a$ = acceleration due to gravity = $-g$ (negative because it acts downward, opposite to the initial velocity)
  • $t$ = time
  So, the equation becomes: $v = u - gt$

• Nature of the v-t Graph: Since $v = u - gt$ is a linear equation in $t$, the velocity-time graph is a straight line with:
  • A positive intercept on the velocity axis (equal to $u$ at $t = 0$)
  • A negative slope (equal to $-g$), indicating decreasing velocity over time

• Velocity at Maximum Height: At the highest point, the velocity becomes zero. This occurs at time: $t = \frac{u}{g}$ After this time, the body begins to fall downward, and the velocity becomes negative (assuming upward as positive).

• Symmetry in Motion: The motion is symmetric. The velocity decreases linearly to zero on the way up and then increases in magnitude (but negative in direction) on the way down.

Thus, the correct $v$–$t$ graph must be a straight line starting at $v = u$, decreasing with constant slope, crossing zero at $t = u/g$, and continuing into negative values.

--- Step-by-Step Derivation:

Step 1: Define the coordinate system

Let’s take upward direction as positive. At $t = 0$, the body is thrown upwards with initial velocity $u$.

Step 2: Write the velocity equation

Under constant acceleration $a = -g$, the velocity at time $t$ is: $v(t) = u - gt$

Step 3: Analyze the graph characteristics

• The graph of $v$ vs $t$ is a straight line.
• At $t = 0$, $v = u$ → positive intercept on velocity axis.
• The slope of the line is $-g$, which is negative and constant.
• The line crosses the time axis (i.e., $v = 0$) at $t = \frac{u}{g}$.
• For $t > \frac{u}{g}$, $v$ becomes negative, indicating downward motion.

Step 4: Compare with given options

Let’s analyze each option:

• Option A: Shows a curve (not a straight line). Velocity decreases non-linearly. This is incorrect because acceleration is constant, so $v$–$t$ must be linear.
• Option B: Shows a straight line starting from $v = u$, but the slope is positive. This would imply acceleration in the same direction as velocity, which is wrong.
• Option C: Shows a straight line starting from $v = 0$ and increasing. This would represent a body starting from rest and accelerating upward — not applicable here.
• Option D: Shows a straight line starting from a positive velocity $u$, decreasing with constant negative slope, crossing zero, and going negative. This perfectly matches the derived equation $v = u - gt$.

Conclusion: Option D correctly represents the velocity-time graph for a body thrown vertically upwards.

--- Common Traps & Exam Tip:

Students often make the following mistakes:

• Confusing acceleration direction: Many assume acceleration is positive because the body is moving upward. But acceleration due to gravity always acts downward, so it must be negative in the upward-positive coordinate system.
• Misinterpreting the v-t graph shape: Some expect a curved graph, thinking acceleration changes. But since $g$ is constant, the graph must be a straight line.
• Ignoring the zero crossing: Forgetting that velocity becomes zero at maximum height and then negative on the way down. The graph must cross the time axis.
• Choosing Option A due to misconception: Option A looks like a "slowing down" curve, which some associate with deceleration, but it’s not linear — hence incorrect.

Exam Tip: Always remember:

• For constant acceleration, $v$–$t$ graph is a straight line.
• Slope of $v$–$t$ graph = acceleration.
• If upward is positive, acceleration due to gravity is negative → negative slope.
• Initial velocity is positive → graph starts above zero.

So, the correct graph must be a straight line with negative slope, starting at $v = u$, crossing zero, and continuing downward. That’s Option D.