One-Dimensional Motion: Displacement and Average Velocity Problems

Theoretical Framework of One-Dimensional Kinematics

  • Position and Coordinate Systems:

    • The position of a material point (particle) moving along a straight line is specified by its coordinate xx relative to a chosen origin OO at a given time instant tt.
    • Coordinates can have positive (x>0x > 0), zero (x=0x = 0), or negative (x<0x < 0) values depending on the chosen orientation of the coordinate axis.

  • Displacement versus Distance:

    • Displacement (Δx\Delta x): A vector quantity defined as the change in position between two instants of time, t1t_1 and t2t_2:     Δx=x2−x1\Delta x = x_2 - x_1
    • Displacement depends exclusively on the initial and final positions, independent of the actual path taken. It carries both magnitude and sign (direction along the coordinate axis).
    • Distance Traveled (dd or ss): A scalar quantity representing the total length of the actual path covered by the particle during the motion. It is always non-negative (d≥0d \ge 0).

  • Average Velocity versus Average Speed:

    • Average Velocity (vavgv_{\text{avg}} or vmv_m): The ratio of the displacement to the time interval during which that displacement occurred:     vavg=ΔxΔt=x2−x1t2−t1v_{\text{avg}} = \frac{\Delta x}{\Delta t} = \frac{x_2 - x_1}{t_2 - t_1}
    • Average velocity is a directed (vector) quantity; its algebraic sign matches that of the displacement Δx\Delta x.
    • Average Speed (savgs_{\text{avg}} or vrrugore, mesv_{\text{rrugore, mes}}: The ratio of the total distance traveled to the total elapsed time interval:     savg=dtotalΔttotals_{\text{avg}} = \frac{d_{\text{total}}}{\Delta t_{\text{total}}}
    • Average speed is strictly a scalar quantity and is always greater than or equal to the magnitude of the average velocity (savg≥∣vavg∣s_{\text{avg}} \ge |v_{\text{avg}}|).

Problem Set: Displacement and Average Velocity in One Dimension

Textbook page showing problems on one-dimensional motion and average velocity

  • Problem 1: Runner's Out-and-Back Motion

    • Problem Description: A runner traverses a path of 2 km2\,\text{km} in 5 min5\,\text{min} and then walks for 10 min10\,\text{min} to return to the starting point.
    • Sub-problem (a) — Average speed for the first 5 min5\,\text{min}:
    • Path traveled: d1=2 kmd_1 = 2\,\text{km}
    • Time elapsed: t1=5 min=560 h=112 ht_1 = 5\,\text{min} = \frac{5}{60}\,\text{h} = \frac{1}{12}\,\text{h}
    • Calculation:       savg,1=d1t1=2 km112 h=24 km/hs_{\text{avg}, 1} = \frac{d_1}{t_1} = \frac{2\,\text{km}}{\frac{1}{12}\,\text{h}} = 24\,\text{km/h}
    • In SI units:       savg,1=2000 m300 s=203 m/s≈6.67 m/ss_{\text{avg}, 1} = \frac{2000\,\text{m}}{300\,\text{s}} = \frac{20}{3}\,\text{m/s} \approx 6.67\,\text{m/s}
    • Sub-problem (b) — Average speed during the return interval:
    • Path traveled in return: d2=2 kmd_2 = 2\,\text{km}
    • Time elapsed: t2=10 min=1060 h=16 ht_2 = 10\,\text{min} = \frac{10}{60}\,\text{h} = \frac{1}{6}\,\text{h}
    • Calculation:       savg,2=d2t2=2 km16 h=12 km/hs_{\text{avg}, 2} = \frac{d_2}{t_2} = \frac{2\,\text{km}}{\frac{1}{6}\,\text{h}} = 12\,\text{km/h}
    • In SI units:       savg,2=2000 m600 s=103 m/s≈3.33 m/ss_{\text{avg}, 2} = \frac{2000\,\text{m}}{600\,\text{s}} = \frac{10}{3}\,\text{m/s} \approx 3.33\,\text{m/s}
    • Sub-problem (c) — Average speed for the entire duration:
    • Total distance covered: dtotal=d1+d2=2 km+2 km=4 kmd_{\text{total}} = d_1 + d_2 = 2\,\text{km} + 2\,\text{km} = 4\,\text{km}
    • Total time elapsed: ttotal=t1+t2=5 min+10 min=15 min=0.25 ht_{\text{total}} = t_1 + t_2 = 5\,\text{min} + 10\,\text{min} = 15\,\text{min} = 0.25\,\text{h}
    • Calculation:       savg, total=dtotalttotal=4 km0.25 h=16 km/hs_{\text{avg, total}} = \frac{d_{\text{total}}}{t_{\text{total}}} = \frac{4\,\text{km}}{0.25\,\text{h}} = 16\,\text{km/h}
    • In SI units:       savg, total=4000 m900 s=409 m/s≈4.44 m/ss_{\text{avg, total}} = \frac{4000\,\text{m}}{900\,\text{s}} = \frac{40}{9}\,\text{m/s} \approx 4.44\,\text{m/s}
    • Sub-problem (d) — Overall average velocity:
    • Because the runner returns to the exact initial point, the final position equals the initial position: xf=xix_f = x_i.
    • Total displacement:       Δx=xf−xi=0 km\Delta x = x_f - x_i = 0\,\text{km}
    • Calculation:       vavg=Δxttotal=0 km0.25 h=0 km/hv_{\text{avg}} = \frac{\Delta x}{t_{\text{total}}} = \frac{0\,\text{km}}{0.25\,\text{h}} = 0\,\text{km/h}

  • Problem 2: Runner with Partial Return

    • Problem Description: Solve Problem 1 under the condition that during the return leg, the runner walks only halfway along the path (12×2 km=1 km\frac{1}{2} \times 2\,\text{km} = 1\,\text{km}) for 10 min10\,\text{min} and then stops.
    • Parameters:
    • Outward stage: d1=2 kmd_1 = 2\,\text{km}, t1=5 mint_1 = 5\,\text{min}, displacement Δx1=+2 km\Delta x_1 = +2\,\text{km}
    • Return stage: d2=1 kmd_2 = 1\,\text{km}, t2=10 mint_2 = 10\,\text{min}, displacement Δx2=−1 km\Delta x_2 = -1\,\text{km}
    • Total time: ttotal=5 min+10 min=15 min=0.25 ht_{\text{total}} = 5\,\text{min} + 10\,\text{min} = 15\,\text{min} = 0.25\,\text{h}
    • Total distance traveled: dtotal=2 km+1 km=3 kmd_{\text{total}} = 2\,\text{km} + 1\,\text{km} = 3\,\text{km}
    • Net displacement: Δx=Δx1+Δx2=+2 km−1 km=+1 km\Delta x = \Delta x_1 + \Delta x_2 = +2\,\text{km} - 1\,\text{km} = +1\,\text{km}
    • Calculated Values:
    • First 5 min5\,\text{min} average speed: savg,1=24 km/hs_{\text{avg}, 1} = 24\,\text{km/h}
    • Return interval average speed:       savg,2=1 km16 h=6 km/hs_{\text{avg}, 2} = \frac{1\,\text{km}}{\frac{1}{6}\,\text{h}} = 6\,\text{km/h}
    • Entire time average speed:       savg, total=3 km0.25 h=12 km/hs_{\text{avg, total}} = \frac{3\,\text{km}}{0.25\,\text{h}} = 12\,\text{km/h}
    • Entire time average velocity:       vavg=Δxttotal=+1 km0.25 h=+4 km/hv_{\text{avg}} = \frac{\Delta x}{t_{\text{total}}} = \frac{+1\,\text{km}}{0.25\,\text{h}} = +4\,\text{km/h}

  • Problem 3: One-Dimensional Coordinate Displacements of a Material Point

    • Problem Description: A material point occupies the following positions along an axis:
    • Coordinate x0=+5 mx_0 = +5\,\text{m} at time instant t0=0 st_0 = 0\,\text{s}
    • Coordinate x1=−7 mx_1 = -7\,\text{m} at time instant t1=6 st_1 = 6\,\text{s}
    • Coordinate x2=+2 mx_2 = +2\,\text{m} at time instant t2=10 st_2 = 10\,\text{s}
    • Sub-problem (a) — Interval from t=0 st = 0\,\text{s} to t=6 st = 6\,\text{s}:
    • Displacement:       Δx01=x1−x0=−7 m−(+5 m)=−12 m\Delta x_{01} = x_1 - x_0 = -7\,\text{m} - (+5\,\text{m}) = -12\,\text{m}
    • Time interval:       Δt01=6 s−0 s=6 s\Delta t_{01} = 6\,\text{s} - 0\,\text{s} = 6\,\text{s}
    • Average velocity:       vavg,a=Δx01Δt01=−12 m6 s=−2 m/sv_{\text{avg}, a} = \frac{\Delta x_{01}}{\Delta t_{01}} = \frac{-12\,\text{m}}{6\,\text{s}} = -2\,\text{m/s}
    • Sub-problem (b) — Interval from t=6 st = 6\,\text{s} to t=10 st = 10\,\text{s}:
    • Displacement:       Δx12=x2−x1=+2 m−(−7 m)=+9 m\Delta x_{12} = x_2 - x_1 = +2\,\text{m} - (-7\,\text{m}) = +9\,\text{m}
    • Time interval:       Δt12=10 s−6 s=4 s\Delta t_{12} = 10\,\text{s} - 6\,\text{s} = 4\,\text{s}
    • Average velocity:       vavg,b=Δx12Δt12=+9 m4 s=+2.25 m/sv_{\text{avg}, b} = \frac{\Delta x_{12}}{\Delta t_{12}} = \frac{+9\,\text{m}}{4\,\text{s}} = +2.25\,\text{m/s}
    • Sub-problem (c) — Interval from t=0 st = 0\,\text{s} to t=10 st = 10\,\text{s}:
    • Displacement:       Δx02=x2−x0=+2 m−(+5 m)=−3 m\Delta x_{02} = x_2 - x_0 = +2\,\text{m} - (+5\,\text{m}) = -3\,\text{m}
    • Time interval:       Δt02=10 s−0 s=10 s\Delta t_{02} = 10\,\text{s} - 0\,\text{s} = 10\,\text{s}
    • Average velocity:       vavg,c=Δx02Δt02=−3 m10 s=−0.3 m/sv_{\text{avg}, c} = \frac{\Delta x_{02}}{\Delta t_{02}} = \frac{-3\,\text{m}}{10\,\text{s}} = -0.3\,\text{m/s}

  • Problem 4: Multi-Segment Journey of a Motorist

    • Problem Description: A motorist initiates a trip of 200 km200\,\text{km} at 120012^{00}.
    • Sub-problem (a) — Non-stop journey arriving at 173017^{30}:
    • Departure time: 120012^{00}; Arrival time: 173017^{30}
    • Elapsed travel time: Δt=5.5 h\Delta t = 5.5\,\text{h}
    • Total distance: d=200 kmd = 200\,\text{km}
    • Calculation:       savg=200 km5.5 h=40011 km/h≈36.36 km/hs_{\text{avg}} = \frac{200\,\text{km}}{5.5\,\text{h}} = \frac{400}{11}\,\text{km/h} \approx 36.36\,\text{km/h}
    • Sub-problem (b) — Trip with an intermediate rest stop:
    • The motorist drives for 3 h3\,\text{h}, pauses for 0.5 h0.5\,\text{h}, and resumes driving, still arriving at 173017^{30}.
    • Total time elapsed from start to finish remains unchanged:       Δttotal=17.5 h−12.0 h=5.5 h\Delta t_{\text{total}} = 17.5\,\text{h} - 12.0\,\text{h} = 5.5\,\text{h}
    • Net distance covered remains d=200 kmd = 200\,\text{km}.
    • Overall average speed across the whole travel window:       savg=200 km5.5 h≈36.36 km/hs_{\text{avg}} = \frac{200\,\text{km}}{5.5\,\text{h}} \approx 36.36\,\text{km/h}
    • (Note: If calculated strictly over active moving time Δtmove=5.5 h−0.5 h=5.0 h\Delta t_{\text{move}} = 5.5\,\text{h} - 0.5\,\text{h} = 5.0\,\text{h}, the moving average speed would be 200 km5 h=40 km/h\frac{200\,\text{km}}{5\,\text{h}} = 40\,\text{km/h}, but the standard definition across the trip duration uses Δttotal\Delta t_{\text{total}}).
    • Sub-problem (c) — Round-trip average speed:
    • The motorist rests at the destination for 2 h2\,\text{h}, then travels back home over an additional 6 h6\,\text{h}.
    • Total distance for round-trip: dround=200 km+200 km=400 kmd_{\text{round}} = 200\,\text{km} + 200\,\text{km} = 400\,\text{km}
    • Total elapsed time:       Δtround=5.5 h+2 h+6 h=13.5 h\Delta t_{\text{round}} = 5.5\,\text{h} + 2\,\text{h} + 6\,\text{h} = 13.5\,\text{h}
    • Calculation:       savg, round=400 km13.5 h=80027 km/h≈29.63 km/hs_{\text{avg, round}} = \frac{400\,\text{km}}{13.5\,\text{h}} = \frac{800}{27}\,\text{km/h} \approx 29.63\,\text{km/h}
    • Sub-problem (d) — Net displacement of the motorist:
    • Because the motorist returns home to the starting position:       Δxtotal=xfinal−xinitial=0 km\Delta x_{\text{total}} = x_{\text{final}} - x_{\text{initial}} = 0\,\text{km}

  • Problem 5: Rectilinear Travel at Variable Speeds

    • Problem Description: A motorist travels along a straight road at an average speed of 80 km/h80\,\text{km/h} for 2.5 h2.5\,\text{h}, and subsequently at 40 km/h40\,\text{km/h} for 1.5 h1.5\,\text{h}.
    • Sub-problem (a) — Total displacement for the 4 h4\,\text{h} trip:
    • Segment 1 displacement:       Δx1=v1×t1=80 km/h×2.5 h=200 km\Delta x_1 = v_1 \times t_1 = 80\,\text{km/h} \times 2.5\,\text{h} = 200\,\text{km}
    • Segment 2 displacement:       Δx2=v2×t2=40 km/h×1.5 h=60 km\Delta x_2 = v_2 \times t_2 = 40\,\text{km/h} \times 1.5\,\text{h} = 60\,\text{km}
    • Total displacement:       Δxtotal=Δx1+Δx2=200 km+60 km=260 km\Delta x_{\text{total}} = \Delta x_1 + \Delta x_2 = 200\,\text{km} + 60\,\text{km} = 260\,\text{km}
    • Sub-problem (b) — Average speed for the entire trip:
    • Total elapsed time:       ttotal=t1+t2=2.5 h+1.5 h=4 ht_{\text{total}} = t_1 + t_2 = 2.5\,\text{h} + 1.5\,\text{h} = 4\,\text{h}
    • Calculation:       savg=dtotalttotal=260 km4 h=65 km/hs_{\text{avg}} = \frac{d_{\text{total}}}{t_{\text{total}}} = \frac{260\,\text{km}}{4\,\text{h}} = 65\,\text{km/h}

  • Problem 6: Highway Odometer and Reversal Analysis

    • Problem Description: Entering a highway, a driver observes the kilometer post at 520 km520\,\text{km}. Traveling the full highway down to kilometer post 0 km0\,\text{km} takes 6 h6\,\text{h}. At this point, wanting to make a short stop, he turns back for 40 km40\,\text{km}, driving for 30 min30\,\text{min}.
    • Sub-problem (a) — Average speed for the entire 560 km560\,\text{km} trip in km/h\text{km/h}:
    • Total distance traveled:       dtotal=520 km+40 km=560 kmd_{\text{total}} = 520\,\text{km} + 40\,\text{km} = 560\,\text{km}
    • Total time elapsed:       ttotal=6 h+30 min=6 h+0.5 h=6.5 ht_{\text{total}} = 6\,\text{h} + 30\,\text{min} = 6\,\text{h} + 0.5\,\text{h} = 6.5\,\text{h}
    • Calculation:       savg=560 km6.5 h=112013 km/h≈86.15 km/hs_{\text{avg}} = \frac{560\,\text{km}}{6.5\,\text{h}} = \frac{1120}{13}\,\text{km/h} \approx 86.15\,\text{km/h}
    • Sub-problem (b) — Average velocity for the entire trip:
    • Initial position coordinate: xi=520 kmx_i = 520\,\text{km}
    • Intermediate position: x=0 kmx = 0\,\text{km}
    • Final position coordinate after turning back 40 km40\,\text{km}: xf=40 kmx_f = 40\,\text{km}
    • Net displacement:       Δx=xf−xi=40 km−520 km=−480 km\Delta x = x_f - x_i = 40\,\text{km} - 520\,\text{km} = -480\,\text{km}
    • Calculation:       vavg=Δxttotal=−480 km6.5 h=−96013 km/h≈−73.85 km/hv_{\text{avg}} = \frac{\Delta x}{t_{\text{total}}} = \frac{-480\,\text{km}}{6.5\,\text{h}} = -\frac{960}{13}\,\text{km/h} \approx -73.85\,\text{km/h}
    • Sub-problem (c) — Average speed for the final 40 km40\,\text{km}:
    • Distance: d2=40 kmd_2 = 40\,\text{km}
    • Time: t2=30 min=0.5 ht_2 = 30\,\text{min} = 0.5\,\text{h}
    • Calculation:       savg,2=40 km0.5 h=80 km/hs_{\text{avg}, 2} = \frac{40\,\text{km}}{0.5\,\text{h}} = 80\,\text{km/h}

  • Problem 7: Supersonic and Subsonic Transatlantic Flight Comparison

    • Problem Description:
    • Ocean width: D=5500 kmD = 5500\,\text{km}
    • Reference speed of sound in air: standard value vsound≈340 m/s=1224 km/hv_{\text{sound}} \approx 340\,\text{m/s} = 1224\,\text{km/h}
    • Supersonic aircraft speed: vsuper=2.4×vsoundv_{\text{super}} = 2.4 \times v_{\text{sound}}vsuper=2.4×1224 km/h=2937.6 km/hv_{\text{super}} = 2.4 \times 1224\,\text{km/h} = 2937.6\,\text{km/h}
    • Subsonic aircraft speed: vsub=0.9×vsoundv_{\text{sub}} = 0.9 \times v_{\text{sound}}vsub=0.9×1224 km/h=1101.6 km/hv_{\text{sub}} = 0.9 \times 1224\,\text{km/h} = 1101.6\,\text{km/h}
    • Sub-problem (a) — Flight time for supersonic aircraft:
    • Expression:       tflight, super=Dvsuper=5500 km2937.6 km/h≈1.872 h≈1 h 52 mint_{\text{flight, super}} = \frac{D}{v_{\text{super}}} = \frac{5500\,\text{km}}{2937.6\,\text{km/h}} \approx 1.872\,\text{h} \approx 1\,\text{h}\,52\,\text{min}
    • Sub-problem (b) — Flight time for subsonic aircraft:
    • Expression:       tflight, sub=Dvsub=5500 km1101.6 km/h≈4.993 h≈5.0 ht_{\text{flight, sub}} = \frac{D}{v_{\text{sub}}} = \frac{5500\,\text{km}}{1101.6\,\text{km/h}} \approx 4.993\,\text{h} \approx 5.0\,\text{h}
    • Sub-problems (c) & (d) — Overall door-to-door average speeds:
    • Additional ground and terminal overhead: 2 h2\,\text{h} at origin (city-to-airport transit and luggage drop/check-in) plus 2 h2\,\text{h} at destination (luggage claim and airport-to-city transit), giving a combined overhead of:       toverhead=2 h+2 h=4 ht_{\text{overhead}} = 2\,\text{h} + 2\,\text{h} = 4\,\text{h}
    • (c) Supersonic door-to-door average speed:
      • Total door-to-door time:         ttotal, super=1.872 h+4.0 h=5.872 ht_{\text{total, super}} = 1.872\,\text{h} + 4.0\,\text{h} = 5.872\,\text{h}
      • Door-to-door average speed:         savg, super=5500 km5.872 h≈936.6 km/hs_{\text{avg, super}} = \frac{5500\,\text{km}}{5.872\,\text{h}} \approx 936.6\,\text{km/h}
    • (d) Subsonic door-to-door average speed:
      • Total door-to-door time:         ttotal, sub=4.993 h+4.0 h=8.993 ht_{\text{total, sub}} = 4.993\,\text{h} + 4.0\,\text{h} = 8.993\,\text{h}
      • Door-to-door average speed:         savg, sub=5500 km8.993 h≈611.6 km/hs_{\text{avg, sub}} = \frac{5500\,\text{km}}{8.993\,\text{h}} \approx 611.6\,\text{km/h}

  • Problem 8: Experimental Determination of Speed Following Instrument Failure

    • Problem Context: While traveling along a highway, a vehicle's speedometer fails to function.
    • Procedural Methodology for Determining Speed:
    • The driver measures the time interval Δt\Delta t required to traverse a known fixed distance Δx\Delta x marked by consecutive highway kilometer posts (e.g., 1 km1\,\text{km} markers or 100 m100\,\text{m} reflector posts).
    • The vehicle's speed is computed directly via the definition of uniform motion velocity:       v=ΔxΔtv = \frac{\Delta x}{\Delta t}
    • For instance, if an interval of Δx=1 km\Delta x = 1\,\text{km} is traversed in a measured time of Δt=36 s\Delta t = 36\,\text{s}, the speed is:       v=1000 m36 s=27.78 m/s=100 km/hv = \frac{1000\,\text{m}}{36\,\text{s}} = 27.78\,\text{m/s} = 100\,\text{km/h}