A quadcopter takes off vertically from ground level
PHYSICS · Multi-part · A Level · Physics · Mechanics and materials · Mechanics and materials
Problem
A quadcopter takes off vertically from ground level, ascends, hovers, and then descends back to the ground. Its altitude (vertical position relative to the ground) as a function of time is described below.
Segment I (0 s ≤ t ≤ 4 s): The altitude increases smoothly from 0 m to 20 m while the upward velocity decreases linearly from 10 m/s to 0 m/s.
Segment II (4 s ≤ t ≤ 6 s): The altitude remains constant at 20 m.
Segment III (6 s ≤ t ≤ 8 s): The motors are shut off; the quadcopter begins to fall freely under a constant downward acceleration of magnitude 10 m/s², starting from rest at t = 6 s and altitude 20 m. It reaches the ground (altitude 0 m) exactly at t = 8 s.
Ignore air resistance in Segment III. Answer all parts qualitatively and quantitatively as requested.
(a) Construct a carefully labeled velocity–time graph for the entire 0-to-8-s interval. Include numerical values at the endpoints of each segment.
(b) On a separate set of axes, sketch the corresponding acceleration–time graph, indicating the sign and magnitude of acceleration in each segment.
(c) Using your velocity graph, calculate the total amount of time the quadcopter’s altitude exceeds 10 m.
(d) Demonstrate quantitatively, using an area or slope argument, that the motion in Segment III is consistent with a constant downward acceleration of 10 m/s² given the starting conditions at t = 6 s and the touchdown at t = 8 s.
(e) Explain how the differences between Segment I (ascent) and Segment II (hover) are reflected in both your velocity and acceleration graphs.