A student builds a drone that ascends vertically at 4 m/s while expanding a rectangular grid of sensors covering a volume of 1,440 m³ per minute. If the grid width increases at 2 m/s and height at 3 m/s, what is the current width of the grid?

["Title: How a Student’s Drone Expands a Sensor Grid — Solving the Math Behind Vertical Ascent and 3D Sensor Coverage", "---", "Introduction\nIn a fascinating blend of aerospace engineering and sensor technology, a student has designed a unique drone capable of ascending vertically at 4 meters per second while simultaneously expanding a rectangular grid of environmental sensors through a dynamic 3D space. At first glance, the system would seem to focus solely on vertical movement, but the drone’s sensor array actively increases in volume—covering 1,440 cubic meters per minute. Understanding how the width of this expanding grid relates to the drone’s motion requires a careful analysis of its vertical lift and horizontal widening.", "In this article, we decode the physics and geometry behind this innovative drone’s sensor deployment, solving for the current width of the rectangular grid covering an expanding volume.", "---", "The Physics of Vertical Ascent and Volume Expansion", "The drone ascends vertically at a constant speed of 4 meters per second. In one minute (60 seconds), it rises:\n[\n\ ext{Vertical distance} = 4, \ ext{m/s} \ imes 60, \ ext{s} = 240, \ ext{m}\n]", "During this same minute, the sensor grid expands to cover a total volume of 1,440 m³ — meaning the volume swept per minute is:\n[\n\ ext{Volume per minute} = 1,440, \ ext{m}^3\n]", "The drone spreads its rectangular sensor grid horizontally and vertically, forming a vertical rectangular prism-shaped coverage volume. However, the grid’s width is increasing over time while the drone ascends, meaning the grid’s cross-section grows from a smaller size to a larger one over 60 seconds.", "Let’s define:\n- ( w ) = current width of the grid (in meters) — the starting width at t = 0\n- ( w + \Delta w ) = width after 60 seconds\n- Vertical speed = 4 m/s\n- Grid height grows at 3 m/s (consistent with vertical motion and grid expansion)", "Since both the width and height expand simultaneously, the full 3D expansion volume per minute reflects the changing cross-sectional area times vertical speed.", "But crucially, the total volume of sensor coverage per minute is the volume swept by the moving grid:", "[\n\ ext{Volume} = (\ ext{initial width} \ imes \ ext{height increase}) \ imes \ ext{vertical distance}\n]", "But more accurately: since the volume increases linearly with time and the drone ascends uniformly, the gradually increasing grid forms a tapered 3D expansion current a rectangular shape extending upward.", "However, simplifying based on the constant rate and uniform expansion, the rate of volume increase equals:\n[\n\frac{dV}{dt} = \ ext{rate at which area expands vertically times vertical speed}\n]", "The volume swept in one minute is generated by the rectangle’s area growing over vertical travel:\n[\n\frac{dV}{dt} = (\ ext{width}(t) \cdot \ ext{height}(t)) \ imes \ ext{vertical speed}\n]", "But because both width and height change linearly over 60 seconds, and volume increases at 1,440 m³ per minute, we use:\n[\n\frac{dV}{dt} = 1,440, \ ext{m}^3/\ ext{min} = \frac{1,440}{60} = 24, \ ext{m}^3/\ ext{s}\n]", "So,\n[\n(\ ext{area at time } t) \ imes (\ ext{vertical speed}) = 24\n]", "At ( t = 60, \ ext{s} ), vertical distance = 240 m, and total volume covered = 1,440 m³, so:\n[\n(\ ext{current width}) \ imes (\ ext{height at } t=60) \ imes 4, \ ext{m/s} \cdot 60, \ ext{s} = 1,440, \ ext{m}^3\n]", "Wait—simpler:\nThe rate of volume increase is directly tied to the cross-sectional area times vertical speed. Since the grid expands vertically at 3 m/s and horizontally at 2 m/s (width increases at 2 m/s), the area at any moment is:", "[\nA(t) = w(t) \cdot h(t)\n]", "Given:\n- ( \frac{dh}{dt} = 3, \ ext{m/s} )\n- ( \frac{dw}{dt} = 2, \ ext{m/s} )\n- Initial width ( w(0) = w ) (unknown)\n- At ( t = 60 ) s, rate of volume increase:\n[\n\frac{dV}{dt} = A(t) \cdot \frac{dh}{dt} = w(t) \cdot h(t) \cdot 3 = 24, \ ext{m}^3/\ ext{s}\n]", "But since width and height grow linearly, ( w(t) = w + 2t ), ( h(t) = h_0 + 3t ). Initially, at ( t = 0 ), the width is ( w ), and we don’t know ( h_0 ), but we can define ( h_0 ) in terms of the expansion.", "However, the problem implies the sensor grid expands — so at ( t = 0 ), the width is minimal, increasing at 2 m/s. The full volume covered per minute is generated by the continuous expansion:", "But a clearer interpretation: the drone sweeps a volume of 1,440 m³ per minute as it ascends and expands a rectangular grid. Since vertical speed is constant and expansion is linear, the average width × average height × 4 m/s = 1,440.", "Let’s assume that over the 60-second interval, the width increases from ( w ) to ( w + 2 \ imes 60 = w + 120 ), and height increases from ( h ) to ( h + 3 \ imes 60 = h + 180 ). But initial conditions aren’t given.", "Alternatively, the volume swept per minute is the area under the expanding rectangle moving upward — but since both width and height grow linearly during the minute, the average area times speed equals 1,440.", "A more elegant and likely intended model: the drone’s sensor grid sweeps a vertical rectangular annular column (like a prism) whose base width increases linearly and height increases uniformly.", "But the simplest and most consistent approach:\nTotal volume per minute = (initial width × vertical height gain) × vertical speed × time? No — incorrect.", "Better definition: The volume covered per minute is ( \ ext{width} \ imes \ ext{height} \ imes \ ext{vertical speed} ), but width and height are changing.", "Instead, if the sensor grid expands such that its horizontal extent (width) increases at 2 m/s and vertical height at 3 m/s, then the rate at which volume increases is:", "[\n\frac{dV}{dt} = (\ ext{width}(t) \cdot \ ext{height}(t)) \cdot \frac{dh}{dt}\n]", "But ( \frac{dh}{dt} = 3, \ ext{m/s} ), and ( \frac{dw}{dt} = 2, \ ext{m/s} ). Let initial width = ( w ), initial height = ( h_0 ). Then:", "[\n\frac{dV}{dt} = w(t) \cdot h(t) \cdot 3 = (w + 2t)(h_0 + 3t) \cdot 3\n]", "Total volume over 60 seconds:", "[\nV = \int_0^{60} 3(w + 2t)(h_0 + 3t), dt = 1,440\n]", "But we have two unknowns.", "However, note: the student’s drone builds a sensor grid expanding at a constant rate — likely meaning width and height grow linearly from zero? But that would give zero volume.", "Alternatively, the drone starts at a small sensor footprint and expands outward. But the rate of volume increase is given as 1,440 m³ per minute at this moment, and we are to find the current width.", "Wait — perhaps the rate of volume increase is determined by current dimensions: expanding 2 m/s in width and 3 m/s in height, but vertical speed is separate.", "But the vertical speed affects how fast area is swept.", "Actually, the correct model is:\nThe drone’s sensor array forms a rectangular prism whose height grows at 3 m/s (so at t = 60, height = 180 m), and whose width grows at 2 m/s (so current width = ?), and vertical speed is 4 m/s. The total volume swept per minute by this expanding prism, moving upward, is formed by integrating the cross-sectional area over the ascent.", "But the cross-sectional area at any moment is width × height, and vertical speed scales the time, so:", "The total volume swept in 60 seconds is:\n[\nV = \int_0^{60} \left[ w(t) \cdot h(t) \right] \cdot \frac{dh}{dt} , dt\n= 3 \int_0^{60} w(t) h(t) , dt\n]", "Assume linear growth:\n- ( w(t) = w + 2t )\n- ( h(t) = h_0 + 3t )\nBut we don’t know ( h_0 ). However, at ( t = 0 ), the drone begins its mission — but the sensor grid must be initialized. Likely, the "expansion" starts from t=0, so perhaps ( h_0 = 0 )? But width increases at 2 m/s, so at t=0, width = 0? Contradiction.", "Alternatively, “expanding a rectangular grid” implies the grid has a fixed starting width — but the problem states width increases at 2 m/s, so it must start from zero. But then area and volume would be zero.", "Ah — here’s the key: the rate of expansion is given as width increases at 2 m/s and height at 3 m/s — this describes the current rates at the moment we’re analyzing. So at ( t = 0 ), width is increasing at 2 m/s, height at 3 m/s.", "But the volume swept per minute is determined by the area at that instant multiplied by vertical speed and time, but only if the expansion rate is instantaneous.", "But volume flow rate is:\n[\n\frac{dV}{dt} = (\ ext{width} \ imes \ ext{height}) \ imes \frac{dh}{dt}\n]", "Wait — that’s not correct. Volume is being swept as a moving slice: as the drone ascends, each vertical slice at height ( z ) subtends a rectangle of area ( w(z) \ imes h_{\ ext{total}} ), but ( h_{\ ext{total}} = 3 \ imes 60 = 180, \ ext{m} ) at the end of the minute.", "But actually, the total volume covered is the integral over depth of the cross-sectional area at each elevation.", "But since the drone moves uniformly, and"]