Velocity–time graphs Transitional Reasoning Practice

Physics · Motion · Physics · Describing Motion

9 Physics practice questions on Velocity–time graphs. Questions that build towards applied reasoning.

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Questions

  1. 1.

    A motorized lift platform moves vertically along a straight track inside a test facility. Its velocity–time log reveals three successive stages over a 16 s interval: it accelerates from rest at t = 0 s to +8 m/s at t = 4 s, maintains a constant velocity of +8 m/s until t = 10 s, and then undergoes a constant acceleration until its velocity reaches -4 m/s at t = 16 s. What is the platform's displacement during the period in which its speed is decreasing, and what is its net displacement over the entire 16 s test?

    • A) +12 m and +84 m
    • B) +12 m and +76 m
    • C) +16 m and +76 m
    • D) +16 m and +84 m
  2. 2.

    An automated research carriage moves along a straight calibrated track. Its velocity–time graph shows that from t=0 st = 0\text{ s} to t=6 st = 6\text{ s}, its velocity changes linearly from −12 m/s-12\text{ m/s} to +6 m/s+6\text{ m/s}. From t=6 st = 6\text{ s} to t=10 st = 10\text{ s}, it continues with a different constant acceleration such that its displacement over this four-second interval is +40 m+40\text{ m}. What are the carriage's acceleration during the first interval and its final velocity at t=10 st = 10\text{ s}?

    • A) Acceleration: +1 m/s2+1\text{ m/s}^2; Final velocity: +10 m/s+10\text{ m/s}
    • B) Acceleration: +3 m/s2+3\text{ m/s}^2; Final velocity: +14 m/s+14\text{ m/s}
    • C) Acceleration: +3 m/s2+3\text{ m/s}^2; Final velocity: +8 m/s+8\text{ m/s}
    • D) Acceleration: −1 m/s2-1\text{ m/s}^2; Final velocity: +14 m/s+14\text{ m/s}
  3. 3.

    An automated rail inspection cart begins from rest at the origin (x = 0 m) at t = 0 s and travels along a straight track. Its telemetry log records that it accelerates uniformly to +10 m/s at t = 5 s, decelerates uniformly to come momentarily to rest at t = 8 s, accelerates in the reverse direction to -4 m/s at t = 12 s, and finally decelerates uniformly until stopping at t = 14 s. At what time does the cart reach its maximum distance in the positive direction from the origin, and what is its position at that instant?

    • A) t = 5 s and x = +25 m
    • B) t = 8 s and x = +15 m
    • C) t = 14 s and x = +28 m
    • D) t = 8 s and x = +40 m
  4. 4.

    A mobile robotic platform starts from rest at the origin (x=0 mx = 0\text{ m}) at t=0 st = 0\text{ s} and travels along a straight track. Its velocity–time log shows that it accelerates uniformly to +8 m/s+8\text{ m/s} at t=2 st = 2\text{ s}, maintains this constant velocity until t=4 st = 4\text{ s}, and then decelerates uniformly to come momentarily to rest at t=6 st = 6\text{ s}. Immediately after, it accelerates uniformly in the reverse direction to −4 m/s-4\text{ m/s} at t=8 st = 8\text{ s} and continues at this constant velocity of −4 m/s-4\text{ m/s} thereafter. At what time tt does the platform return to its starting position, and what total distance has it traveled at that moment?

    • A) t=12 st = 12\text{ s} and total distance = 48 m48\text{ m}
    • B) t=14 st = 14\text{ s} and total distance = 56 m56\text{ m}
    • C) t=15 st = 15\text{ s} and total distance = 64 m64\text{ m}
    • D) t=16 st = 16\text{ s} and total distance = 68 m68\text{ m}
  5. 5.

    A diagnostic sensor records the motion of an automated shuttle along a straight calibration track over a 10 s interval. The shuttle's velocity–time graph is a single straight line that starts at -6 m/s at t = 0 s, crosses 0 m/s at t = 4 s, and reaches +9 m/s at t = 10 s. Based on this velocity–time graph, what are the total distance traveled by the shuttle and its average velocity over the entire 10 s test?

    • A) Total distance = 15 m, Average velocity = +3.9 m/s
    • B) Total distance = 27 m, Average velocity = +2.7 m/s
    • C) Total distance = 15 m, Average velocity = +1.5 m/s
    • D) Total distance = 39 m, Average velocity = +1.5 m/s
  6. 6.

    A telemetry sensor on a straight test track logs the velocity of an electric transport cart over a 5 s period. The cart has an initial velocity of 4 m/s at t = 0 s, accelerates uniformly at +2 m/s² for 3 s to reach a peak velocity of 10 m/s, and then immediately decelerates uniformly at a constant rate until coming to rest at t = 5 s. Based on this velocity–time data, what are the total displacement of the cart and its average acceleration over the entire 5 s interval?

    • A) 25 m and -0.8 m/s²
    • B) 31 m and -1.5 m/s²
    • C) 31 m and -0.8 m/s²
    • D) 35 m and +0.8 m/s²
  7. 7.

    An autonomous inspection robot travels along a straight track, starting from rest at 0 s and accelerating uniformly to a steady velocity vv at 4 s. It maintains this constant velocity until 10 s, after which it decelerates uniformly to come to a complete stop at 12 s. If the total distance covered during this 12 s journey is 72 m, what are the steady velocity vv and the magnitude of the deceleration during the final 2 s?

    • A) 6 m/s and 3 m/s²
    • B) 12 m/s and 6 m/s²
    • C) 8 m/s and 2 m/s²
    • D) 8 m/s and 4 m/s²
  8. 8.

    Two automated sorting carriers, X and Y, move along parallel straight tracks in a warehouse, both starting from the same baseline at t=0 st = 0\text{ s}. Carrier X travels at a constant velocity of 6 m/s6\text{ m/s} throughout an 8 s8\text{ s} test. Carrier Y starts from rest and accelerates uniformly to 12 m/s12\text{ m/s} at t=4 st = 4\text{ s}, then continues at that constant velocity until t=8 st = 8\text{ s}. Based on their velocity–time graphs, at what time does Carrier Y catch up to Carrier X, and what is their displacement from the baseline at that moment?

    • A) At t=2 st = 2\text{ s}, with a displacement of 12 m12\text{ m}
    • B) At t=4 st = 4\text{ s}, with a displacement of 24 m24\text{ m}
    • C) At t=6 st = 6\text{ s}, with a displacement of 36 m36\text{ m}
    • D) At t=8 st = 8\text{ s}, with a displacement of 48 m48\text{ m}
  9. 9.

    A sensor-guided test cart moves along a straight track, and its motion is tracked over a 12-second interval. Starting from rest at the origin at t=0 st = 0\text{ s}, the cart accelerates uniformly to a velocity of +6 m/s+6\text{ m/s} at t=4 st = 4\text{ s}, then uniformly decelerates to 0 m/s0\text{ m/s} at t=7 st = 7\text{ s} and reaches −2 m/s-2\text{ m/s} at t=8 st = 8\text{ s}, after which it maintains a constant velocity of −2 m/s-2\text{ m/s} until t=12 st = 12\text{ s}. Based on this velocity–time data, which statement correctly describes the cart's motion?

    • A) The cart's acceleration is zero at t=7 st = 7\text{ s} because its instantaneous velocity is zero.
    • B) The cart changes its direction of motion at t=4 st = 4\text{ s} when its speed begins to decrease.
    • C) The total distance traveled by the cart throughout the 12-second test is equal to 12 m12\text{ m}.
    • D) The cart is farthest from its starting position at t=7 st = 7\text{ s}, reaching a displacement of +21 m+21\text{ m}.

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