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Graph slopes and areas

Physics • One-Dimensional Motion • Transitional Reasoning • 13 questions

Graph slopes and areas Transitional Reasoning Download PDF

Physics · Motion · Physics · One-Dimensional Motion

13 Physics practice questions on Graph slopes and areas. Questions that build towards applied reasoning.

Questions

  1. 1.

    A remote-controlled probe travels along a straight track, starting from the origin at t=0 st = 0\text{ s}. Its velocity–time graph starts at v=+16 m/sv = +16\text{ m/s} and maintains a constant slope of −4 m/s2-4\text{ m/s}^2 until t=6 st = 6\text{ s}. From t=6 st = 6\text{ s} onward, the slope of the velocity–time graph is zero, so the probe continues at constant velocity. Based on the graphical meanings of slope and bounded area, at what time does the probe return to the origin, and what total distance has it traveled up to that moment?

    • A) t=8 st = 8\text{ s} and total distance = 64 m64\text{ m}
    • B) t=7 st = 7\text{ s} and total distance = 48 m48\text{ m}
    • C) t=9 st = 9\text{ s} and total distance = 32 m32\text{ m}
    • D) t=9 st = 9\text{ s} and total distance = 64 m64\text{ m}
  2. 2.

    A test shuttle on a straight track starts from rest at t = 0 s and undergoes a uniform acceleration phase lasting 4 s. Immediately afterward, it enters a braking phase where the negative slope of its velocity–time graph has twice the magnitude of the initial acceleration slope, bringing the shuttle smoothly to a stop. Given that the total area bounded by the velocity–time graph and the time axis is 72 m, what are the shuttle's peak velocity and the total duration of its motion?

    • A) Peak velocity = 18 m/s; total duration = 8 s
    • B) Peak velocity = 12 m/s; total duration = 12 s
    • C) Peak velocity = 24 m/s; total duration = 6 s
    • D) Peak velocity = 36 m/s; total duration = 6 s
  3. 3.

    A sensor-instrumented cart travels along a straight track with an initial velocity of +10 m/s+10\text{ m/s} at t=0 st = 0\text{ s}. Telemetry records indicate that the area bounded by its acceleration–time graph and the time axis from t=0 st = 0\text{ s} to t=5 st = 5\text{ s} is −25 m/s-25\text{ m/s}, followed by a constant acceleration of +4.0 m/s2+4.0\text{ m/s}^2 from t=5 st = 5\text{ s} to t=9 st = 9\text{ s}. Based on the graphical meanings of bounded area and slope, what is the cart's velocity at t=9 st = 9\text{ s}, and what is the slope of the secant line representing its average acceleration on the velocity–time graph over the entire 9 s9\text{ s} interval?

    • A) −9.0 m/s-9.0\text{ m/s} and −1.0 m/s2-1.0\text{ m/s}^2
    • B) +1.0 m/s+1.0\text{ m/s} and −1.0 m/s2-1.0\text{ m/s}^2
    • C) +1.0 m/s+1.0\text{ m/s} and −0.5 m/s2-0.5\text{ m/s}^2
    • D) −15.0 m/s-15.0\text{ m/s} and −2.8 m/s2-2.8\text{ m/s}^2
  4. 4.

    A glider moves along a straight air track with an initial velocity of -8 m/s at t = 0 s. Its acceleration–time graph consists of two constant segments: an acceleration of +2 m/s² from t = 0 s to t = 6 s, followed by an acceleration of +4 m/s² from t = 6 s to t = 9 s. Based on the relationships between graph slopes and bounded areas, at what time does the tangent to the glider's position–time graph become horizontal, and what is the slope of the position–time graph at t = 9 s?

    • A) t = 4 s and +16 m/s
    • B) t = 4 s and +24 m/s
    • C) t = 2 s and +16 m/s
    • D) t = 3 s and +12 m/s
  5. 5.

    An automated rover moves along a straight test track, starting from rest at t=0 st = 0\text{ s}. Its velocity–time graph shows a constant positive slope of 2.5 m/s22.5\text{ m/s}^2 until t=4 st = 4\text{ s}, remains horizontal until t=10 st = 10\text{ s}, and then maintains a constant negative slope of −5.0 m/s2-5.0\text{ m/s}^2 until the rover comes to a complete stop. Based on the slope and area relationships of the velocity–time graph, what is the rover's average velocity over the entire duration of the motion?

    • A) 5.0 m/s5.0\text{ m/s}
    • B) 6.0 m/s6.0\text{ m/s}
    • C) 6.5 m/s6.5\text{ m/s}
    • D) 7.5 m/s7.5\text{ m/s}
  6. 6.

    Two automated transport pods, Pod X and Pod Y, travel along parallel straight tracks, departing simultaneously from the same starting line at t=0 st = 0\text{ s}. Pod X travels with a constant velocity such that its velocity–time graph is a horizontal line at v=12 m/sv = 12\text{ m/s}, while Pod Y starts from rest and exhibits a velocity–time graph with a constant slope of 3 m/s23\text{ m/s}^2. Based on the graphical meanings of slope and bounded area, at what time does Pod Y match the velocity of Pod X, and at what subsequent time does Pod Y overtake Pod X?

    • A) Pod Y matches the velocity of Pod X at t=4 st = 4\text{ s} and overtakes Pod X at t=6 st = 6\text{ s}.
    • B) Pod Y matches the velocity of Pod X at t=4 st = 4\text{ s} and overtakes Pod X at t=8 st = 8\text{ s}.
    • C) Pod Y matches the velocity of Pod X at t=2 st = 2\text{ s} and overtakes Pod X at t=4 st = 4\text{ s}.
    • D) Pod Y matches the velocity of Pod X at t=4 st = 4\text{ s} and overtakes Pod X at t=12 st = 12\text{ s}.
  7. 7.

    A sensor-equipped test cart travels along a straight track under constant acceleration. Data analysis reveals that the slope of its position–time graph at t=10 st = 10\text{ s} is 18 m/s18\text{ m/s}, and the bounded area under its velocity–time graph from t=0 st = 0\text{ s} to t=10 st = 10\text{ s} is 120 m120\text{ m}. Based on the graphical meanings of slope and area, what are the cart's initial velocity at t=0 st = 0\text{ s} and the slope of its velocity–time graph?

    • A) v0=12 m/sv_0 = 12\text{ m/s} and slope =0.6 m/s2= 0.6\text{ m/s}^2
    • B) v0=3 m/sv_0 = 3\text{ m/s} and slope =1.5 m/s2= 1.5\text{ m/s}^2
    • C) v0=8 m/sv_0 = 8\text{ m/s} and slope =1.0 m/s2= 1.0\text{ m/s}^2
    • D) v0=6 m/sv_0 = 6\text{ m/s} and slope =1.2 m/s2= 1.2\text{ m/s}^2
  8. 8.

    An automated track inspector moves along a straight rail, beginning at an initial position of x = +15 m at t = 0 s. Telemetry records show that its velocity changes linearly from -6 m/s at t = 0 s to +10 m/s at t = 8 s, and then remains steady at +10 m/s until t = 12 s. Using the slope and area relationships of the velocity–time graph, at what time does the inspector reach its minimum position coordinate, and what is its position at t = 12 s?

    • A) t = 3 s and x = 56 m
    • B) t = 3 s and x = 71 m
    • C) t = 4 s and x = 63 m
    • D) t = 3 s and x = 89 m
  9. 9.

    A battery-powered cart moves along a straight track, starting from the origin at t=0 st = 0\text{ s}. Its velocity–time graph is a single straight line with a constant slope of −3 m/s2-3\text{ m/s}^2 that intersects the time axis at t=4 st = 4\text{ s} and terminates at t=6 st = 6\text{ s}. Based on the graphical relationships of slope and area, what are the total distance traveled by the cart and its net displacement over the entire 6 s6\text{ s} interval?

    • A) Total distance = 24 m; net displacement = +24 m
    • B) Total distance = 18 m; net displacement = +30 m
    • C) Total distance = 36 m; net displacement = +12 m
    • D) Total distance = 30 m; net displacement = +18 m
  10. 10.

    During a performance test along a straight track, a guided transport pod's position–time graph exhibits a constant slope from x=0 mx = 0\text{ m} at t=0 st = 0\text{ s} to x=24 mx = 24\text{ m} at t=4 st = 4\text{ s}. Between t=4 st = 4\text{ s} and t=8 st = 8\text{ s}, the pod undergoes constant acceleration, and the area bounded by its velocity–time graph and the time axis is 40 m40\text{ m}. Based on the fundamental relationships between graph slopes, areas, and kinematic quantities, what are the pod's acceleration during the second interval and its position at t=8 st = 8\text{ s}?

    • A) a=2 m/s2a = 2\text{ m/s}^2 and x=64 mx = 64\text{ m}
    • B) a=1 m/s2a = 1\text{ m/s}^2 and x=64 mx = 64\text{ m}
    • C) a=1 m/s2a = 1\text{ m/s}^2 and x=40 mx = 40\text{ m}
    • D) a=5 m/s2a = 5\text{ m/s}^2 and x=40 mx = 40\text{ m}
  11. 11.

    A linear transport capsule starts from rest at t=0t = 0 s and moves along a straight vacuum tube. Its acceleration–time graph shows a constant acceleration of +4 m/s2+4\text{ m/s}^2 from t=0t = 0 s to t=3t = 3 s, followed by a linear decrease from +4 m/s2+4\text{ m/s}^2 at t=3t = 3 s to −4 m/s2-4\text{ m/s}^2 at t=7t = 7 s. Based on the geometric area and slope relationships of motion graphs, at what time does the capsule achieve its maximum forward velocity, and what is this velocity?

    • A) At t=3t = 3 s, with vmax⁡=12v_{\max} = 12 m/s
    • B) At t=5t = 5 s, with vmax⁡=16v_{\max} = 16 m/s
    • C) At t=7t = 7 s, with vmax⁡=12v_{\max} = 12 m/s
    • D) At t=5t = 5 s, with vmax⁡=20v_{\max} = 20 m/s
  12. 12.

    An autonomous shuttle operates along a straight test track, starting from the origin at t = 0 s. For the first 8 seconds, its position–time graph is a straight line with a constant slope of +15 m/s. From t = 8 s to t = 12 s, the shuttle brakes with constant deceleration, bringing it smoothly to rest at t = 12 s. Based on the relationships between graph slopes, areas, and kinematic motion, what is the shuttle's total displacement from t = 0 s to t = 12 s?

    • A) 180 m
    • B) 90 m
    • C) 120 m
    • D) 150 m
  13. 13.

    An automated laboratory cart travels along a straight track, and its motion is recorded on a velocity–time graph. Between t=0 st = 0\text{ s} and t=6 st = 6\text{ s}, the graph is a line segment that rises uniformly from v=2 m/sv = 2\text{ m/s} to v=8 m/sv = 8\text{ m/s}. From t=6 st = 6\text{ s} to t=10 st = 10\text{ s}, the graph descends linearly from v=8 m/sv = 8\text{ m/s} to v=0 m/sv = 0\text{ m/s}. Based on the graphical relationships for slope and bounded area, what are the cart's acceleration during the first interval and its total displacement over the entire 10 s10\text{ s} trip?

    • A) Acceleration is 1 m/s21\text{ m/s}^2 and total displacement is 50 m50\text{ m}
    • B) Acceleration is 1 m/s21\text{ m/s}^2 and total displacement is 46 m46\text{ m}
    • C) Acceleration is 1.33 m/s21.33\text{ m/s}^2 and total displacement is 46 m46\text{ m}
    • D) Acceleration is 1.33 m/s21.33\text{ m/s}^2 and total displacement is 80 m80\text{ m}