Notes on Velocity and Acceleration: Examples 2.1–2.9 (Kinematics)

Example 2.1: Average and instantaneous velocities

Scenario: A cheetah starts 20 m east of a vehicle (origin at the vehicle). At t = 0 the cheetah runs due east toward an antelope located 50 m east of the vehicle. During the first 2.0 s, the cheetah

$x(t)=20\text{ m}+(5.0\text{ m/s}^2)t^2.$

• This implies a constant acceleration of

  $a(t)=\frac{dv}{dt}=\frac{d}{dt}(\,dx/dt\,)=10\text{ m/s}^2,$
  

  since $v(t)=\frac{dx}{dt}=2(5.0\text{ m/s}^2)t=10t\text{ m/s}.$

• Therefore the instantaneous velocity grows linearly with time: $v(t)=10t\text{ m/s}.$

• (a) Displacement between t1 = 1.0 s and t2 = 2.0 s:
  

  $\Delta x=x(2.0)-x(1.0)=[20+5(2.0)^2]-[20+5(1.0)^2]=(40-25)=15\text{ m}.$

• (b) Average velocity during the interval:
  

  $v_{\text{avg}}=\frac{\Delta x}{\Delta t}=\frac{15\text{ m}}{(2.0-1.0)\text{ s}}=15\text{ m/s}.$

• (c) Instantaneous velocity at t1 = 1.0 s by taking Δt = 0.1 s, 0.01 s, 0.001 s (limit of the average velocity over a vanishing interval):

  • Δt = 0.1 s: $v\approx\frac{x(1.1)-x(1.0)}{0.1}=\frac{26.05-25.0}{0.1}=10.5\text{ m/s}.$

  • Δt = 0.01 s: $v\approx\frac{x(1.01)-x(1.0)}{0.01}=\frac{25.1005-25.0}{0.01}=10.05\text{ m/s}.$

  • Δt = 0.001 s: $v\approx\frac{x(1.001)-x(1.0)}{0.001}=\frac{25.010005-25.0}{0.001}=10.005\text{ m/s}.$

  • As Δt \rightarrow 0, $v(1.0\text{ s})=10.0\text{ m/s}.$

• (d) Expression for instantaneous velocity as a function of time and evaluate at t = 1.0 s and t = 2.0 s:
  

  $v(t)=\frac{dx}{dt}=10t\quad\Rightarrow\quad v(1.0\text{ s})=10\text{ m/s},\ v(2.0\text{ s})=20\text{ m/s}.$

• Since a(t) is constant (10 m/s^2), the instantaneous acceleration is
  

  $a(t)=\frac{dv}{dt}=10\text{ m/s}^2.$

Example 2.2: Average acceleration

• Setup: An astronaut moves along a straight line. The partner measures velocity every 2.0 s starting at t = 1.0 s:
  

  • [1.0, 3.0]: Δv = 1.2-0.8 = 0.4 m/s \rightarrow $a_{\text{avg}}=\frac{0.4}{2.0}=0.20\text{ m/s}^2$; speed increases (v from 0.8 to 1.2, both positive).

  • [3.0, 5.0]: Δv = 1.6-1.2 = 0.4 m/s \rightarrow $a_{\text{avg}}=0.20\text{ m/s}^2$; speed increases.

  • [5.0, 7.0]: Δv = 1.2-1.6 = -0.4 m/s \rightarrow $a_{\text{avg}}=-0.20\text{ m/s}^2$; speed decreases.

  • [7.0, 9.0]: Δv = -0.4-1.2 = -1.6 m/s \rightarrow $a_{\text{avg}}=-0.80\text{ m/s}^2$; speed decreases (velocity crosses zero between these times).

  • [9.0, 11.0]: Δv = -1.0-(-0.4) = -0.6 m/s \rightarrow $a_{\text{avg}}=-0.30\text{ m/s}^2$; speed increases in magnitude (more negative velocity).

  • [11.0, 13.0]: Δv = -1.6-(-1.0) = -0.6 m/s \rightarrow $a_{\text{avg}}=-0.30\text{ m/s}^2$; speed increases.

  • [13.0, 15.0]: Δv = -0.8-(-1.6) = +0.8 m/s \rightarrow $a_{\text{avg}}=+0.40\text{ m/s}^2$; speed decreases.

• (b) Interpretations in terms of speed depend on the sign and magnitude of velocity; note intervals with positive velocity see speed increase, while negative velocity intervals can also increase speed if |v| grows even as v becomes more negative.

Example 2.3: Average and instantaneous accelerations

• Given: The x-velocity of a car is
  

  $U<em>x(t)=60\text{ m/s} + (0.50\text{ m/s}^2) t,$ so the instantaneous x-acceleration is $a</em>x(t)=\frac{dU_x}{dt}=0.50\text{ m/s}^2.$

• (a) Change in x-velocity from t1 = 1.0 s to t2 = 3.0 s:
  

  $\Delta U<em>x = a</em>x\Delta t = (0.50)(3.0-1.0)=1.0\text{ m/s}.$

• (b) Average x-acceleration over this interval:
  

  $a<em>{\text{avg}}=\frac{\Delta U</em>x}{\Delta t}=\frac{1.0}{2.0}=0.50\text{ m/s}^2.$

• (c) Instantaneous x-acceleration at t1 = 1.0 s by refining Δt (0.1, 0.01, 0.001 s): since (a_x(t)) is constant, the approximate values approach 0.50 m/s^2.

• (d) Expression for instantaneous acceleration as a function of time:
  

  $a_x(t)=0.50\text{ m/s}^2.$
  

  Evaluate at t1 = 1.0 s and t = 3.0 s: $a<em>x(1.0\text{ s})=0.50\text{ m/s}^2,\ a</em>x(3.0\text{ s})=0.50\text{ m/s}^2.$

Example 2.4: Constant-acceleration calculations

• Given: A motorcyclist accelerates east with a constant acceleration $a<em>x=4.0\text{ m/s}^2.$ At t = 0, he is 5.0 m east of the signpost and moving east at $v</em>0=15\text{ m/s}.$, his x-coordinate follows

• (a) Position x and velocity v at t = 2.0 s:

  • Velocity: $v(t)=v<em>0+a</em>x t=15+4(2)=23\text{ m/s}.$

  • Position: $x(t)=x<em>0+v</em>0 t + \frac{1}{2} a_x t^2 = 5 + 15(2) + \frac{1}{2}(4)(2)^2 = 5+30+8=43\text{ m}.$

• (b) Where is he when his speed is 25 m/s? Solve for t from $v = v<em>0 + a</em>x t = 25:$
  

  $t=\frac{25-15}{4}=2.5\text{ s}.$
  Then
  

  $x(2.5)=5+15(2.5)+\frac{1}{2}(4)(2.5)^2=5+37.5+12.5=55.0\text{ m}.$

Example 2.5: Two objects with different accelerations

• Setup: A motorist passes a school crossing at 15 m/s (speed limit 10 m/s). At that instant, a police officer starts in pursuit from rest with constant acceleration $a=3.0\text{ m/s}^2.$ (a) How much time until the officer passes the motorist? (b) At that moment, what is the officer

• (a)

  • S speed? (c) How far has each traveled?

• Model with origin at the crossing and t = 0 when the officer starts:

  • Motorist: $x<em>m(t)=v</em>m t=15 t.$

  • Officer: $x_o(t)= \frac{1}{2} a t^2 = \frac{1}{2}(3) t^2 = 1.5 t^2.$

• Solve for catch-up: set $x<em>m = x</em>o:$
  

  $15 t = 1.5 t^2 \Rightarrow t(15-1.5 t)=0 \Rightarrow t=0\text{ or } t=10\text{ s}.$

• At t = 10 s,

  • Officer speed: $v_o=a t=3(10)=30\text{ m/s}.$

  • Distances: $x<em>m(10)=15(10)=150\text{ m},\ x</em>o(10)=1.5(100)=150\text{ m}.$

Example 2.6: A freely falling coin

• Setup: A coin is dropped from rest from the Leaning Tower of Pisa (ignore air resistance). Let the positive direction be downward for y.

• Acceleration: $a_y = g = 9.80\text{ m/s}^2.$

• With initial height $y<em>0 = 0$ and initial velocity $v</em>0 = 0$, the position and velocity after time t are:

  • Position: $y(t)=\tfrac{1}{2} g t^2 = \tfrac{1}{2}(9.80) t^2.$

  • Velocity: $v_y(t)=g t = 9.80t\text{ m/s}.$

• Values:

  • At t = 1.0 s: $y=4.9\text{ m},\ v=9.8\text{ m/s}.$

  • At t = 2.0 s: $y=19.6\text{ m},\ v=19.6\text{ m/s}.$

  • At t = 3.0 s: $y=44.1\text{ m},\ v=29.4\text{ m/s}.$

Example 2.7: Up-and-down motion in free fall

• Setup: A ball is thrown vertically upward from a roof level with initial speed 15.0 m/s (upward). Ignore air resistance; acceleration is downward with magnitude $g = 9.80\text{ m/s}^2.$

• Let y be height above the roof railing (y = 0 at the roof). Initial conditions: y(0)=0, $v(0)=+15.0\text{ m/s}$, and $a = -g = -9.80\text{ m/s}^2$ throughout.

• (a) Position and velocity at t = 1.00 s and t = 4.00 s after release:

  • Using $v(t)=v<em>0 - g t;\ y(t)=v</em>0 t - \frac{1}{2} g t^2:$

  • At t = 1.00 s: $v = 15.0 - 9.8(1.0) = 5.2\text{ m/s};\ y = 15.0(1.0) - 0.5(9.8)(1.0)^2 = 15 - 4.9 = 10.1\text{ m}.$

  • At t = 4.00 s: $v = 15.0 - 9.8(4.0) = 15 - 39.2 = -24.2\text{ m/s};\ y = 15(4.0) - 0.5(9.8)(4.0)^2 = 60 - 78.4 = -18.4\text{ m}$ (below roof perspective).

• (b) Ball

  • s velocity when it is 5.00 m above the railing (i.e., y = 5.00 m): solve $v = v_0 - g t$ and $y(t) = 5.$

  • Solve $y(t)=v<em>0 t - \frac{1}{2} g t^2 = 5$ with $v</em>0=15,\ g=9.8:$
    

    $-\tfrac{1}{2}g t^2 + v_0 t - 5 = 0 \Rightarrow -4.9 t^2 + 15 t - 5 = 0.$
    Solutions: $t = \frac{15 \pm \sqrt{15^2 - 4(-4.9)(-5)}}{2\cdot 4.9} = \frac{15 \pm \sqrt{127}}{9.8} \approx 0.381,\ 2.68\text{ s}.$
    Corresponding velocities: for $t \approx 0.381\text{ s},\ v=15 - 9.8(0.381) \approx 11.27\text{ m/s};$
    for $t \approx 2.68\text{ s},\ v=15 - 9.8(2.68) \approx -11.27\text{ m/s}.$\ (Two moments when the ball is 5 m above the railing, once on the way up and once on the way down.)

• (c) Maximum height reached: at $v = 0$, so $t<em>{\text{max}} = v</em>0/g = 15/9.8 \approx 1.53\text{ s;}$ maximum height
  

  $y<em>{\text{max}} = v</em>0^2/(2g) = 15^2/(2\cdot 9.8) \approx 11.45\text{ m}.$

• (d) Acceleration when at maximum height: $a = -g = -9.80\text{ m/s}^2.$

Example 2.8: Two solutions or one?

• Question: At what time after release does the ball in Example 2.7 fall 5.00 m below the roof railing? (i.e., y = -5.00 m if downward is positive or y = -5 depending on your coordinate choice.)

• Using the same upward-positive convention with $y(t)=v<em>0 t - \frac{1}{2} g t^2$ and $y = -5$ (five meters below roof): $-\tfrac{1}{2}g t^2 + v</em>0 t + 5 = 0 \Rightarrow 4.9 t^2 - 15 t - 5 = 0.$

• Solve: $t = \frac{15 \pm \sqrt{15^2 + 4\cdot 4.9 \cdot 5}}{2\cdot 4.9} = \frac{15 \pm \sqrt{225+98}}{9.8} = \frac{15 \pm \sqrt{323}}{9.8}.$

• Numerically: $t \approx \frac{15+17.97}{9.8} \approx 3.36\text{ s},$
  and a negative root (discarded for t > 0). Hence the ball is 5.00 m below the roof at about $t \approx 3.36\text{ s}$ after release.

Example 2.9: Motion with changing acceleration

• Sally

  • s motion along a straight highway starts at t = 0 with x = 50 m and moving in +x with velocity $U_x(0) = 10\text{ m/s}.$ The x-acceleration is
    

  $a_x(t)=2.0\text{ m/s}^2 - (0.10\text{ m/s}^3) t.$

• (a) Find x(t) and $U_x(t):$ integrate acceleration to get velocity and position.

  • Velocity:
    

    $U<em>x(t)=U</em>x(0) + \int<em>0^t a</em>x(\tau)\,d\tau = 10 + \int_0^t \left(2.0 - 0.10\tau\right) d\tau = 10 + 2t - 0.05 t^2.$

  • Position:
    

    $x(t)=x(0) + \int<em>0^t U</em>x(\tau)\,d\tau = 50 + \int_0^t \left(10 + 2\tau - 0.05\tau^2\right) d\tau = 50 + 10t + t^2 - \frac{0.05}{3} t^3 = 50 + 10t + t^2 - \frac{1}{60} t^3.$

• (b) When is the x-velocity greatest? Since $a<em>x(t) = 2.0 - 0.10 t$, the velocity increases while ax > 0, and decreases after $a<em>x = 0$ when t = 20 s. Thus the maximum velocity occurs at $t</em>{\text{max}}=20\text{ s}.$
  The maximum velocity is
  

  $U<em>{x, \text{max}} = U</em>x(20) = 10 + 2(20) - 0.05(20)^2 = 10 + 40 - 20 = 30\text{ m/s}.$

• (c) Value of that maximum velocity: $30\text{ m/s}.$

• (d) Position when the maximum velocity is reached (t = 20 s):
  

  $x(20) = 50 + 10(20) + (20)^2 - \frac{1}{60}(20)^3 = 50 + 200 + 400 - \frac{8000}{60} = 50 + 200 + 400 - 133.333\dots \approx 516.67\text{ m}.$

Key concepts and takeaways

• Instantaneous velocity vs. average velocity:

  • Instantaneous velocity is the slope of the x vs. t curve at a point, i.e., $v(t)=\frac{dx}{dt}.$

  • Average velocity over an interval is the total displacement divided by the time interval: $v_{\text{avg}}=\frac{\Delta x}{\Delta t}.$

• Instantaneous acceleration vs. average acceleration:

  • Instantaneous acceleration is the slope of the v vs. t curve, i.e., $a(t)=\frac{dv}{dt}.$

  • Average acceleration over an interval is the change in velocity divided by the time interval: $a_{\text{avg}}=\frac{\Delta v}{\Delta t}.$

• Constant-acceleration kinematics: when a is constant, the standard relations apply

  • Position: $x(t)=x<em>0+v</em>0 t+\frac{1}{2}a t^2$

  • Velocity: $v(t)=v_0+ a t$

• Basic results for free fall and vertical motion: sign conventions matter; downward acceleration is often taken as negative if upward is positive, etc.

• Real-world relevance: these examples show how to extract kinematic quantities from given time histories, predict future states, and check consistency with units and signs.

참고: LaTeX notation used here follows the guidance: all mathematical expressions are enclosed in double dollar signs ($\dots$). Expressions such as velocities, accelerations, positions, times, and constants (g, m/s, m/s^2, etc.) appear in their standard forms to mirror the problem statements and solutions above.