Reading velocity–time graphs: Class 9 practice
On a velocity–time graph the slope is acceleration and the area under the line is displacement. Read the method, follow the three worked examples, then try the six questions.
What you should know first
- Area of triangle and rectangle
The idea behind reading velocity–time graphs
On a velocity–time graph the slope is acceleration and the area under the line is displacement. A horizontal line means constant velocity; a straight sloping line means uniform acceleration.
Step-by-step method
- Read the velocities at the start and end of the section.
- Slope = Δv ÷ Δt gives acceleration.
- Area (rectangles and triangles) gives displacement.
Worked example 1 (Basic)
A v–t graph rises in a straight line from 0 m/s to 9 m/s in 9 s. Find the acceleration.
- a = (9 − 0)/9
- = 1 m/s²
- Answer: 1 m/s²
Worked example 2 (Standard)
Using the same graph (9 → 24 m/s in 6 s), find the displacement.
- Area = ½(9 + 24) × 6
- = 99 m
- Answer: 99 m
Worked example 3 (Challenge)
A body moves at 24 m/s for 3 s, then decelerates uniformly to rest in 3 s. Find the total distance.
- Rectangle: 24 × 3 = 72 m
- Triangle: ½ × 24 × 3 = 36 m
- Total = 108 m
- Answer: 108 m
Common mistakes and misconceptions
- Reading area as acceleration
- Forgetting the triangle's ½
Practice questions
- (Basic) A v–t graph rises in a straight line from 8 m/s to 23 m/s in 9 s. Find the acceleration.
- (Basic) A v–t graph rises in a straight line from 6 m/s to 17 m/s in 10 s. Find the acceleration.
- (Standard) Using the same graph (9 → 19 m/s in 9 s), find the displacement.
- (Standard) Using the same graph (2 → 17 m/s in 7 s), find the displacement.
- (Challenge) A body moves at 10 m/s for 7 s, then decelerates uniformly to rest in 10 s. Find the total distance.
- (Challenge) A body moves at 11 m/s for 3 s, then decelerates uniformly to rest in 4 s. Find the total distance.
Answer key with full working
- Q1: a = (23 − 8)/9 → = 1.67 m/s² → Answer: 1.67 m/s²
- Q2: a = (17 − 6)/10 → = 1.1 m/s² → Answer: 1.1 m/s²
- Q3: Area = ½(9 + 19) × 9 → = 126 m → Answer: 126 m
- Q4: Area = ½(2 + 17) × 7 → = 66.5 m → Answer: 66.5 m
- Q5: Rectangle: 10 × 7 = 70 m → Triangle: ½ × 10 × 10 = 50 m → Total = 120 m → Answer: 120 m
- Q6: Rectangle: 11 × 3 = 33 m → Triangle: ½ × 11 × 4 = 22 m → Total = 55 m → Answer: 55 m
Syllabus note
This skill is listed under "Motion" for Class 9 in the NCERT/CBSE books. ICSE, IGCSE, IB and state boards may teach it in a different chapter or year.
Answers on this page are generated and automatically re-checked by a second, independent calculation. They have not been reviewed by a teacher — if you spot a problem, tell us on WhatsApp.
Related practice and notes
Frequently asked questions
Which chapter is reading velocity–time graphs in?
"Motion", Class 9 Physics (NCERT/CBSE). Other boards may place it differently.
How were the answers checked?
Answers on this page are generated and automatically re-checked by a second, independent calculation. They have not been reviewed by a teacher — if you spot a problem, tell us on WhatsApp.