Mechanics

Chapter 2: Kinematics in One Dimension

Evan Luo

6 min read

1. Describing Motion with Graphs

Kinematics = describing how things move (no forces yet).

Instead of long sentences, we can let graphs tell the story: a quantity on the vertical axis vs. time on the horizontal axis.

Graph typeWhat its slope givesWhat its area gives
Position-timeVelocity—
Velocity-timeAccelerationDisplacement
Acceleration-time—Change in velocity

If a line is straight, the quantity (velocity or acceleration) is constant. Curved lines mean the quantity is changing.

2. Position-Time Graphs

Think of the graph as a "diary" of where the object is each second.

  • Straight slanted line → uniform motion (same speed every second). Steeper = faster, downward = moving in the negative direction.
  • Curved line → accelerating (speed changing). Instantaneous velocity is the slope of the tangent (the little line that just kisses the curve).

3. Velocity-Time Graphs

This one tells us how fast and in what direction.

  • Flat (horizontal) line = constant velocity.
  • Slanted line = constant acceleration (the slope is aa).
  • Area under the curve between two times gives the displacement in that interval. Negative area means the object moved the opposite way.

4. Acceleration-Time Graphs

Shows how quickly velocity itself is changing.

  • Flat line = constant acceleration.
  • Curved or stepped line = acceleration changing.
  • Area under the curve = change in velocity Δv\Delta v.

5. Constant-Acceleration Motion: The "Big 4" Equations

When the acceleration stays the same, position and velocity change in highly predictable ways (notice the mix of linear and quadratic terms):

vf=vi+aΔtsf=si+viΔt+12a(Δt)2vf 2=vi 2+2a(sf−si)sf=si+12(vi+vf)Δt\begin{aligned} v_f &= v_i + a\Delta t \\[4pt] s_f &= s_i + v_i\Delta t + \tfrac{1}{2}a(\Delta t)^2 \\[4pt] v_f^{\,2} &= v_i^{\,2} + 2a(s_f - s_i) \\[4pt] s_f &= s_i + \tfrac{1}{2}(v_i + v_f)\Delta t \end{aligned}

Symbols: ss = position (can use xx or yy), vv = velocity, aa = acceleration, Δt\Delta t = time interval.

6. Free Fall

Drop it → it accelerates downward at g=9.80 m/s2g = 9.80\ \text{m/s}^2.

Whether thrown up or dropped, while the object is in the air gravity is the only acceleration (ignore air resistance for now). So we can plug a=−ga = -g into any of the 4 equations.

7. Motion on an Incline

Slide a block down a smooth ramp tilted by an angle θ\theta: only part of gravity acts along the slope.

aslope=gsin⁡θa_{\text{slope}} = g\sin\theta

Bigger angle → larger component → faster slide. Axes are rotated to run parallel and perpendicular to the ramp for easier math.

8. Calculus View (Why Slopes & Areas Work)

For curved graphs we use very small slices:

  • Derivative (slope of a tangent) gives the instantaneous rate of change
    • v=dsdtv = \dfrac{ds}{dt}
    • a=dvdta = \dfrac{dv}{dt}
  • Integral (tiny rectangles summed up) gives accumulated change
    • Δs=∫v dt\Delta s = \int v\,dt
    • Δv=∫a dt\Delta v = \int a\,dt

If calculus feels abstract, remember: it's just the ultra-zoomed-in version of "rise over run" and "area under the curve".

9. Key Takeaways

  • Graphs turn motion into easy-to-read pictures.
  • Slope ↔ rate of change; area ↔ accumulated change.
  • Uniform motion → straight lines; curved lines → acceleration.
  • Constant acceleration lets us use the "Big 4" equations.
  • Free-fall and ramp problems are just special cases of constant-aa.

Keep these relations handy, and the rest of mechanics will feel much friendlier!

10. Worked Examples

Example 1: Train and Car at a Crossing

Problem Statement:
A train is moving north at 30 m/s30\ \mathrm{m/s} and decelerating at 0.80 m/s20.80\ \mathrm{m/s}^2 and is 60 m60\ \mathrm{m} from a road crossing. At the same instant, a car is driving east on the road at 20 m/s20\ \mathrm{m/s} and is 45 m45\ \mathrm{m} from the crossing. After a 0.50 s0.50\ \mathrm{s} reaction time, the driver of the car decides to accelerate so as to reach the tracks just as the train arrives. What minimum acceleration does the car need?

Solution Steps:

  1. Time for train to reach crossing
    Use s=vit+12at2s = v_i t + \tfrac12 a t^2:

    60=30 t−0.40 t260 = 30\,t - 0.40\,t^2

    Solve the quadratic −0.40t2+30t−60=0-0.40t^2 + 30t - 60 = 0 to get

    ttrain≈2.06 s.t_{\text{train}} \approx 2.06\ \mathrm{s}.

  2. Distance car travels during reaction

    dreact=vcar treact=20×0.50=10 m.d_{\text{react}} = v_{\text{car}}\,t_{\text{react}} = 20 \times 0.50 = 10\ \mathrm{m}.

    Remaining distance to crossing:

    sremain=45−10=35 m.s_{\text{remain}} = 45 - 10 = 35\ \mathrm{m}.

  3. Time available for acceleration

    tacc=ttrain−treact≈2.06−0.50=1.56 s.t_{\text{acc}} = t_{\text{train}} - t_{\text{react}} \approx 2.06 - 0.50 = 1.56\ \mathrm{s}.

  4. Solve for required acceleration
    Use s=vit+12at2s = v_i t + \tfrac12 a t^2 for the acceleration phase:

    35=20 (1.56)+12a (1.56)2⟹a≈3.12 m/s2.35 = 20\,(1.56) + \tfrac12 a\,(1.56)^2 \quad\Longrightarrow\quad a \approx 3.12\ \mathrm{m/s}^2.

Answer: amin⁡≈3.12 m/s2\displaystyle a_{\min} \approx 3.12\ \mathrm{m/s}^2.

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Example 2: Maximum Height of a Two-Stage Rocket

Problem Statement:
During the first stage of a two-stage rocket, the rocket starts from rest and experiences a constant upward acceleration of 3.50 m/s23.50\ \mathrm{m/s}^2 for 25.0 s25.0\ \mathrm{s}. The second stage then fires for an additional 10.0 s10.0\ \mathrm{s}, increasing the rocket's velocity to a final value of 132.5 m/s132.5\ \mathrm{m/s} upwards. After burnout, the rocket is in free fall.

  • (a) What is the maximum height above the launchpad?
  • (b) How much time passes from launch until maximum height?

Solution Steps:

  1. Stage 1:

    • Initial: y0=0y_0 = 0, v0=0v_0 = 0
    • Acceleration: a1=3.50 m/s2a_1 = 3.50\ \mathrm{m/s}^2, Δt1=25.0 s\Delta t_1 = 25.0\ \mathrm{s}
    • Final velocity: v1=v0+a1 Δt1=0+(3.50)(25.0)=87.5 m/sv_1 = v_0 + a_1\,\Delta t_1 = 0 + (3.50)(25.0) = 87.5\ \mathrm{m/s}
    • Height gained: s1=12(v0+v1) Δt1=12(0+87.5)(25.0)=1093.75 ms_1 = \frac{1}{2}(v_0 + v_1)\,\Delta t_1 = \frac{1}{2}(0 + 87.5)(25.0) = 1093.75\ \mathrm{m}

  2. Stage 2:

    • Initial: y1=s1=1093.75 my_1 = s_1 = 1093.75\ \mathrm{m}, v1=87.5 m/sv_1 = 87.5\ \mathrm{m/s}
    • Final velocity: v2=132.5 m/sv_2 = 132.5\ \mathrm{m/s}, Δt2=10.0 s\Delta t_2 = 10.0\ \mathrm{s}
    • Acceleration: a2=v2−v1Δt2=132.5−87.510.0=4.50 m/s2a_2 = \frac{v_2 - v_1}{\Delta t_2} = \frac{132.5 - 87.5}{10.0} = 4.50\ \mathrm{m/s}^2
    • Height gained: s2=12(v1+v2) Δt2=12(87.5+132.5)(10.0)=1100 ms_2 = \frac{1}{2}(v_1 + v_2)\,\Delta t_2 = \frac{1}{2}(87.5 + 132.5)(10.0) = 1100\ \mathrm{m}
    • Total so far: y2=y1+s2=1093.75+1100=2193.75 my_2 = y_1 + s_2 = 1093.75 + 1100 = 2193.75\ \mathrm{m}

  3. Free Fall (Coast to Apex):

    • Initial: y2=2193.75 my_2 = 2193.75\ \mathrm{m}, v2=132.5 m/sv_2 = 132.5\ \mathrm{m/s}
    • Acceleration: a3=−g=−9.80 m/s2a_3 = -g = -9.80\ \mathrm{m/s}^2
    • Additional height: Δy3=−v222a3=−(132.5)22(−9.80)=896.5 m\Delta y_3 = -\frac{v_2^2}{2 a_3} = -\frac{(132.5)^2}{2(-9.80)} = 896.5\ \mathrm{m}
    • Maximum height: ymax⁡=y2+Δy3=2193.75+896.5=3090.25≈3090 my_{\max} = y_2 + \Delta y_3 = 2193.75 + 896.5 = 3090.25 \approx 3090\ \mathrm{m}
    • Time to coast: Δt3=−v2a3=−132.5−9.80=13.52 s\Delta t_3 = -\frac{v_2}{a_3} = -\frac{132.5}{-9.80} = 13.52\ \mathrm{s}

Answers:

  • Hmax⁡≈3090 mH_{\max} \approx 3090\ \mathrm{m}
  • Ttotal=Δt1+Δt2+Δt3=25.0+10.0+13.52=48.52 sT_{\text{total}} = \Delta t_1 + \Delta t_2 + \Delta t_3 = 25.0 + 10.0 + 13.52 = 48.52\ \mathrm{s}
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