Sky-diver's displacement under gravity and drag approaching terminal velocity

Steps:

• At t=0, no air resistance, so acceleration = g, displacement s ∝ t² (parabolic increase).
• As speed v rises, drag force F_d = kv (or kv²) grows, reducing net force and acceleration.
• When F_d = mg, net force = 0, terminal velocity v_t reached, acceleration = 0.
• Thus, s vs t graph: initially curved upward (increasing slope), then linear (constant slope = v_t).

Why B is correct:

• B shows initial parabolic rise transitioning to straight line, matching s = (1/2)gt² early then s = v_t t after terminal velocity per Newton's second law.

Why the others are wrong:

• A: Linear throughout, implies constant velocity from start, ignores initial free-fall acceleration.
• C: Concave down always, suggests decelerating motion, contradicts increasing then constant speed.
• D: Exponential or irregular curve, doesn't reflect quadratic-to-linear transition to terminal velocity.

Final answer: B