For A, acceleration = 1 m s⁻² and displacement in 5 s = 12.5 m. For B, acceleration = 0.3 m s⁻² and displacement in 10 s = 15 m. Their velocity-time graphs are straight lines from the origin.
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For the first 120 s, displacement = 6 × 120 = 720 m. During the next 6 s, displacement = 6 × 6 + ½ × 1 × 6² = 54 m. Total displacement = 774 m.
The distance travelled is approximately equal to the area under the velocity-time graph. Estimating the areas of the trapeziums from the graph gives a total distance of approximately 45 km.
For constant velocity, the area from 20–100 s is a rectangle: 3 × 80 = 240 m. For decreasing velocity, area from 100–120 s = ½(3+2)×20 = 50 m. Total displacement = 320 m; average acceleration = (2−0)/120 = 0.0167 ...
The perpendicular from the centre bisects the chord, forming two right triangles. Applying the Baudhāyana–Pythagoras Theorem gives the half-chord as √(r² − d²). Hence, the chord length is 2√(r² − d²).