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.
Two cars A and B start moving with a constant acceleration from rest, in a straight line. Car A attains a velocity of 5 m s⁻¹ in 5 s. Car B attains a velocity of 3 m s⁻¹ in 10 s. Plot the velocity-time graphs for both the cars in the same graph. Using the graph, calculate the displacement in the two time intervals mentioned (Hint: Calculate the acceleration in both cases). Then calculate their velocities at five instants of time to plot the graph.
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For car A, acceleration = (5−0)/5 = 1 m s⁻². Its displacement in 5 s = ½ × 1 × 5² = 12.5 m. For car B, acceleration = (3−0)/10 = 0.3 m s⁻². Its displacement in 10 s = ½ × 0.3 × 10² = 15 m. Thus, both velocity-time graphs are straight lines starting from the origin, with slopes 1 and 0.3 m s⁻², respectively.
For more NCERT Solutions of Class 9 Science Exploration Chapter 4 Describing Motion Around Us Question Answer (2026-27)
https://www.tiwariacademy.com/ncert-solutions/class-9/science/exploration-chapter-4/