The following plots show variation of velocity (v), with time (t), of a ball thrown vertically upward, and falling back. Which of the following plots is/are correct?
Show Hint
A velocity-time graph for any object under constant acceleration must be a straight line. If the graph "bounces" back to the positive side (like plot B), it represents a Speed-time graph, not a Velocity-time graph.
Step 1: Understanding the Topic:
This question deals with "Kinematics," specifically motion under gravity. It tests the graphical representation of vector quantities. Understanding how sign conventions apply to velocity and acceleration is crucial for interpreting these graphs. Step 2: Key Formulas and Approach:
Using the first equation of motion:
\[ v = u + at \]
For a ball thrown upwards (taking upward as the positive direction):
$u$ is positive initial velocity.
$a = -g$ (constant downward acceleration).
$v = u - gt$.
Step 3: Detailed Explanation:
Equation form: The relation $v = -gt + u$ is a linear equation of the form $y = mx + c$.
Slope: The slope of a velocity-time graph is acceleration. Since gravity is constant and acts downwards, the slope must be constant and negative.
Sequence of motion: $t = 0$: Velocity is at its maximum positive value.
Moving Up: Velocity decreases linearly.
Peak: Velocity reaches exactly zero.
Falling Down: Velocity increases in the negative direction (downwards).
Graph Analysis: Plot A: Shows constant velocity (Wrong).
Plot B: Velocity becomes positive again after reaching zero. This is a speed-time graph, not velocity (Wrong).
Plot C: A single straight line with a constant negative slope crossing the time axis. This perfectly matches $v = u - gt$.
