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20
Divide by 3.6.
Class 12 · Bangla Version · Physics
Measurement, motion and forces
Use units, motion graphs and resultant forces to solve a mechanics problem.
Shared cross-level core and stretch unit. Start with the core questions, then use the stretch example for deeper work. The same unit is available in related class and board pathways; it is not a full grade-specific or examination-specification course. A teacher should select the appropriate depth and further topics.
Learn the idea
Speed is distance per time. Velocity includes direction; acceleration is change in velocity per time. The gradient of a displacement-time graph is velocity. The area under a velocity-time graph gives displacement with signs. Newton’s F = ma uses resultant force and inertial mass.
Worked example
A trolley changes from 2 to 10 m/s in 4 s at constant acceleration: a = (10−2)/4 = 2 m/s². For mass 3 kg, resultant F = 6 N. Its displacement is average velocity × time = 6 × 4 = 24 m.
- Convert to consistent units.
- Identify whether acceleration is constant.
- Check directions, equations and units.
Stretch example
In uniform circular motion, radial acceleration v²/r points toward the centre even though speed is constant. At v = 6 m/s and r = 3 m, a = 12 m/s².
Your written practice
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3
(15−0)/5.
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8
F = 2 × 4.
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Direction can change, as in uniform circular motion.
Velocity includes direction.
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Displacement in the chosen negative direction.
Explain the sign convention.
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24
F = mv²/r = 2 × 36/3.
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The area is a trapezium with parallel sides u and v and width t, giving average velocity times time.
State constant acceleration and use the graph area.
Motion solution
Create and solve a two-stage straight-line motion problem.
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- Label the time intervals.
- Use a velocity-time sketch.
- Distinguish displacement from distance.
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