Question 8
The speed-time graph for a car is shown in Fig. below
(a) Find how far does the car travel in the first 4 seconds. Shade the area on the graph that represents the distance travelled by the car during the period.
(b) Which part of the graph represents uniform motion of the car?
(a) Distance travelled by the car in the first 4 seconds = area under the slope of the speed-time graph.
To find this area, we will use the square counting method.
On time axis (i.e., x axis) :
5 squares = 2 units
∴ 1 square = units
On speed axis (i.e., y axis) :
3 squares = 2 units
∴ 1 square = units
Area of each square = x = sq units.
We have fully filled squares, more than half filled squares, half filled squares and less than half filled squares in this area.
Area of fully filled square = Area of 1 square = sq units
Area of more than half filled square = Area of 1 square = sq units
Area of half filled square = x Area of 1 square = = sq units
Area of less than half filled square = 0
No. of fully filled squares = 56 [Marked in graph with 'F']
Area of fully filled squares = 56 x = 14.93 sq units
No. of more than half filled squares = 4 [Marked in graph with 'M']
Area of more than half filled squares = 4 x = 1.06 sq units
No. of half filled squares = 3 [Marked in graph with 'H']
Area of half filled squares = 3 x = 0.8 sq units
No. of less than half filled squares = 4 [Marked in graph with 'L']
Area of less than half filled squares = 4 x 0 = 0 sq units
∴ Total Area = 14.93 + 1.06 + 0.4 = 16.39 sq units
∴ Total distance covered in first 4 seconds = Area under speed-time graph = 16.4 m
Hence, the car travels 16.4 m in first 4 seconds
(b) From graph, the speed of the car does not change from the points (x = 6) to (x = 10). Hence, the car is said to be in uniform motion from the 6th to the 10th second.
Chapter 7: Motion — Quick Reference
Quick Revision Points
- Distance (scalar, total path) vs Displacement (vector, shortest path)
- Speed = distance/time (scalar); Velocity = displacement/time (vector)
- Acceleration a = (v−u)/t; unit: m/s²
- Equations: v = u + at; s = ut + ½at²; v² = u² + 2as
- d-t graph slope = speed; v-t graph slope = acceleration; area under v-t = distance
- 1 km/h = 5/18 m/s; Free fall: u = 0, a = g = 9.8 m/s²
- Uniform circular motion: constant speed, changing velocity (direction changes)
- Read the detailed chapter notes for complete coverage of all NCERT topics.
- Practice all NCERT in-text and back exercise questions — they are frequently asked in exams.
- Focus on comparison tables, diagrams, and definitions — these are high-scoring areas.
- For numericals (if applicable), practice at least 20 problems of varying difficulty.
- Refer to the practice question bank (200+ questions) for thorough preparation.
- Detailed Notes: ch07-motion.html
- Practice Questions: 100+ questions with answers in 05-practice-questions/
- Chapter Test: 30-mark test paper in 06-tests/chapter-tests-30marks/
- Formula Sheet: Complete formula reference in 03-teacher-aid/formula-sheet.html