A particle starts at s = 0 with velocity +6 m s⁻¹ and acceleration −2 m s⁻² for 6 s. Construct distance–time, displacement–time, speed–time, velocity–time and acceleration–time graphs, showing the turning point and final values. At t = 1 s a tangent to its displacement graph passes through (0 s, 1 m) and (2 s, 9 m). Use it to find instantaneous velocity, then find average velocity over the full 6 s.

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Turn at 3 s at displacement/distance 9 m. At 6 s distance=18 m, displacement=0, speed=6 m s⁻¹, velocity=−6 m s⁻¹, acceleration=−2 m s⁻². Instantaneous velocity=+4 m s⁻¹; full-trip average velocity=0. v=6−2t crosses zero at t=3; s=6t−t² reaches 9 m there and returns to 0 at 6 s. Draw displacement as a smooth concave-down curve; draw distance matching it until 3 s, then 9+(t−3)² to 18 m. Speed is |6−2t|; velocity is 6−2t; acceleration stays −2. Label all axes and units; mark turning time, endpoints and relevant zero values. Tangent gradient=(9−1)/(2−0)=4 m s⁻¹. Average velocity=(0−0)/6=0.

Construct and interpret all five motion graphs

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