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
Ux(t)=60\text{ m/s} + (0.50\text{ m/s}^2) t, so the instantaneous x-acceleration is ax(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 Ux = ax\Delta t = (0.50)(3.0-1.0)=1.0\text{ m/s}.
(b) Average x-acceleration over this interval:
a{\text{avg}}=\frac{\Delta Ux}{\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: ax(1.0\text{ s})=0.50\text{ m/s}^2,\ ax(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 ax=4.0\text{ m/s}^2. At t = 0, he is 5.0 m east of the signpost and moving east at v0=15\text{ m/s}., his x-coordinate follows
(a) Position x and velocity v at t = 2.0 s:
Velocity: v(t)=v0+ax t=15+4(2)=23\text{ m/s}.
Position: x(t)=x0+v0 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 = v0 + ax t = 25:
t=\frac{25-15}{4}=2.5\text{ s}.
Thenx(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: xm(t)=vm 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 xm = xo:
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: xm(10)=15(10)=150\text{ m},\ xo(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 y0 = 0 and initial velocity v0 = 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)=v0 - g t;\ y(t)=v0 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)=v0 t - \frac{1}{2} g t^2 = 5 with v0=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{\text{max}} = v0/g = 15/9.8 \approx 1.53\text{ s;} maximum height
y{\text{max}} = v0^2/(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)=v0 t - \frac{1}{2} g t^2 and y = -5 (five meters below roof): -\tfrac{1}{2}g t^2 + v0 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:
Ux(t)=Ux(0) + \int0^t ax(\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) + \int0^t Ux(\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 ax(t) = 2.0 - 0.10 t, the velocity increases while ax > 0, and decreases after ax = 0 when t = 20 s. Thus the maximum velocity occurs at t{\text{max}}=20\text{ s}.
The maximum velocity isU{x, \text{max}} = Ux(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)=x0+v0 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.