QPage 48-49: Q4: HC Verma: Concepts of Physics - Volume 1 Chapter-3 Rest and Motion Kinematics Page 48-49: Q4 Solution: A ball is thrown vertically upward with a speed of 20 m/s. Draw …
Students often fail to maintain a consistent sign convention, plotting the downward phase with positive velocity or forgetting that the slope of the $v$-$t$ graph represents acceleration ($-g$), which must remain negative throughout the motion.
A ball is thrown vertically upward with a speed of 20 m/s. Draw a graph showing the velocity of the ball as a function of .....
Multiple Choice Options
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Step-by-Step Verified Solution
Here is the step-by-step solution and the graph description for the motion of a ball thrown vertically upward with an initial speed of 20 m/s.
Step 1: Establish the Sign Convention
- Take upward as positive.
- Take downward as negative.
- Acceleration due to gravity, g = 10 m/s² (downward, so a = –10 m/s²).
Step 2: Determine Key Values
Initial velocity (at t = 0):
u = +20 m/s
Time to reach maximum height (where v = 0):
Using v = u + at
0 = 20 – 10t
⇒ t = 2 seconds
Maximum height:
Using s = ut + ½at²
s = 20(2) + ½(–10)(2)² = 40 – 20 = 20 m
Total time of flight (up and down):
Time up = 2 s, time down = 2 s
⇒ Total time = 4 seconds
Velocity just before hitting the ground:
Using v = u + at with total time t = 4 s
v = 20 – 10(4) = 20 – 40 = –20 m/s (negative sign means downward)
Step 3: Velocity–Time Graph
The graph is a straight line because acceleration is constant.
Time (s) | Velocity (m/s) | Direction |
0 | +20 | Upward |
1 | +10 | Upward |
2 | 0 | (Top) |
3 | –10 | Downward |
4 | –20 | Downward |
Graph: Velocity (v) vs. Time (t)
Features of the graph:
- Straight line with a negative slope (–10 m/s²).
- Slope = acceleration = –10 m/s² (constant).
- Intercept on v-axis = +20 m/s (initial velocity).
- Crosses the time axis at t = 2 s (velocity becomes zero at the highest point).
- The line continues into negative velocities after t = 2 s, indicating downward motion.
- At t = 4 s, velocity is –20 m/s (same speed as initial, but opposite direction).
Step 4: Equation of the Line
The graph represents the equation:
v(t) = 20 – 10t (for 0 ≤ t ≤ 4 s)
Step 5: Key Takeaways
Aspect | Observation |
Shape | Straight line (linear) |
Slope | Negative constant = –10 m/s² (gravity) |
v = 0 at | t = 2 s (maximum height) |
Symmetry | Velocity at t = 0 is +20 m/s; at t = 4 s is –20 m/s. Speeds are equal, directions are opposite. |
Area under the graph | Displacement. Area above time axis = +20 m (upward). Area below = –20 m (downward). Net displacement = 0 (returns to start). |
Final Answer
The velocity–time graph is a straight line with a negative slope (–10 m/s²), starting at +20 m/s at t = 0, crossing zero at t = 2 s, and reaching –20 m/s at t = 4 s. The line is symmetric about the point (2 s, 0), showing that the ball returns to the thrower with the same speed but opposite velocity.