ANAND CLASSES study material and notes to learn how acceleration is determined from the slope of a velocity-time graph with detailed explanations, derivations, MCQs, conceptual questions, worksheets, and test papers for JEE, NEET, and CBSE Class 11 exams.
Derivation of Acceleration from Velocity-Time Graph
Step 1: Definition of Acceleration
Acceleration a is given by: $$a = \frac{\text{Change in velocity}}{\text{Time taken}} = \frac{v_2 – v_1}{t_2 – t_1} \quad \text{(Equation 1)}$$
Step 2: Understanding the Slope of Velocity-Time Graph

From the velocity-time graph, considering the right-angled triangle ABC, we have: $$\tan \theta = \frac{BC}{AC} = \frac{AD}{AC} = \frac{v_2 – v_1}{t_2 – t_1} \quad \text{(Equation 2)}$$
Step 3: Comparing Both Equations
By comparing Equation (1) and Equation (2): $$a = \tan \theta$$
Thus, the acceleration is equal to the slope of the velocity-time graph.
Conceptual Understanding with Example
Consider a car moving with an initial velocity of 5 m/s and increasing its velocity to 25 m/s in 10 seconds. $$a = \frac{25 – 5}{10} = \frac{20}{10} = 2 \text{ m/s}^2$$
The slope of the velocity-time graph for this motion would be 2, which is equal to the acceleration of the car.
Question-Answer Format for JEE, NEET, and CBSE Board Class 11
Q1: How can the acceleration of a particle be determined from a velocity-time graph?
Answer: Acceleration is determined from the slope of the velocity-time graph. If the graph is a straight line, the acceleration is constant and equal to the slope of the line.
Q2: What does a horizontal velocity-time graph indicate?
Answer: A horizontal line in a velocity-time graph represents zero acceleration, meaning the object is moving with constant velocity.
Q3: What happens when the velocity-time graph has a negative slope?
Answer: A negative slope means that the velocity is decreasing with time, indicating negative acceleration (deceleration).
FAQs on Acceleration and Velocity-Time Graph
Q1: What is the unit of acceleration?
Answer: The SI unit of acceleration is m/s².
Q2: Can the velocity-time graph be a curve?
Answer: Yes, if the acceleration is not constant, the velocity-time graph will be a curve.
Q3: How is instantaneous acceleration determined?
Answer: The instantaneous acceleration is the slope of the velocity-time graph at a particular instant.
Multiple Choice Questions (MCQs)
Q1: What does the slope of a velocity-time graph represent?
A) Velocity
B) Acceleration
C) Distance
D) Displacement
Answer: B) Acceleration
Q2: If the velocity-time graph is a straight line parallel to the time axis, what is the acceleration?
A) 0
B) Positive
C) Negative
D) Infinity
Answer: A) 0
Q3: If a velocity-time graph has a negative slope, the motion is:
A) Uniform motion
B) Uniform acceleration
C) Retardation
D) Zero acceleration
Answer: C) Retardation
Do You Know? (Interesting Facts)
- The area under the velocity-time graph represents the displacement of the particle.
- If the acceleration is zero, the velocity-time graph will be a horizontal straight line.
- In free-fall motion, acceleration is 9.8 m/s², which is the acceleration due to gravity.
Worksheet
Question 1:
A train increases its velocity from 30 m/s to 50 m/s in 5 seconds. Find the acceleration.
Question 2:
Draw and explain the velocity-time graph for an object moving with:
- Uniform acceleration
- Non-uniform acceleration
- Constant velocity
Question 3:
If an object comes to rest from a velocity of 40 m/s in 8 seconds, calculate its acceleration.
Test Paper (10 Marks)
Question No. | Question | Marks |
---|---|---|
1 | Define acceleration and its unit. | 2 |
2 | How does the velocity-time graph help in finding acceleration? | 3 |
3 | A body moving with a velocity of 20 m/s accelerates uniformly at 4 m/s² for 5 seconds. Find the final velocity. | 5 |
Quick Revision Points
✔ Acceleration is the rate of change of velocity.
✔ The slope of a velocity-time graph represents acceleration.
✔ If the graph is a straight line, acceleration is constant.
✔ A negative slope indicates deceleration.
✔ The area under the velocity-time graph gives displacement.
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