Comprehensive One-Dimensional Kinematics Study Guide

One-Dimensional Kinematics: Multi-Segment Motion and Average Velocity

• Problem Statement:

  • A car travels in a straight line with an average velocity of $80\,\text{km/h}$ for $2.5\,\text{h}$ and then with an average velocity of $40\,\text{km/h}$ for $1.5\,\text{h}$.

  • $(a)$ What is the total displacement for the $4.0\text{-h}$ trip?

  • $(b)$ What is the average velocity for the total trip?

• Definition of Average Velocity:

  • Average velocity is defined as displacement per unit time:     $v_{avg} = \frac{\Delta x}{\Delta t}$

• Segment-by-Segment Displacement Calculations:

  • First Segment:

  • Average velocity: $v_{avg,1} = 80\,\text{km/h}$

  • Time interval: $\Delta t_1 = 2.5\,\text{h}$

  • Displacement: $\Delta x_1 = v_{avg,1} \times \Delta t_1 = 80\,\text{km/h} \times 2.5\,\text{h} = 200\,\text{km}$

  • Second Segment:

  • Average velocity: $v_{avg,2} = 40\,\text{km/h}$

  • Time interval: $\Delta t_2 = 1.5\,\text{h}$

  • Displacement: $\Delta x_2 = v_{avg,2} \times \Delta t_2 = 40\,\text{km/h} \times 1.5\,\text{h} = 60\,\text{km}$

• Part (a) Solution - Total Displacement:

  • Summing the displacements of both individual segments:     $\Delta x_{total} = \Delta x_1 + \Delta x_2$     $\Delta x_{total} = 200\,\text{km} + 60\,\text{km} = 260\,\text{km}$

• Part (b) Solution - Average Velocity for Total Trip:

  • Total elapsed time for the journey:     $\Delta t_{total} = \Delta t_1 + \Delta t_2 = 2.5\,\text{h} + 1.5\,\text{h} = 4.0\,\text{h}$

  • Calculating average velocity across the entire trip:     $v_{avg,total} = \frac{\Delta x_{total}}{\Delta t_{total}} = \frac{260\,\text{km}}{4.0\,\text{h}} = 65\,\text{km/h}$

Graphical Analysis of Position vs. Time

• Problem Statement:

  • Position as a function of time is plotted on a coordinate grid. Find the average velocities for the time intervals $a$, $b$, $c$, and $d$ indicated on the position-time curve.

• General Formula:

  • Average velocity on a position-time graph corresponds to the slope of the secant line joining the initial and final position points:     $v_{avg} = \frac{x_f - x_i}{t_f - t_i}$

• Interval Analysis and Calculations:

  • Interval $a$ ($t = 0\,\text{s}$ to $t = 4\,\text{s}$):

  • Initial position at $t = 0\,\text{s}$: $x_i = 3\,\text{m}$

  • Final position at $t = 4\,\text{s}$: $x_f = 3\,\text{m}$

  • Average velocity calculation:       $v_{avg,a} = \frac{3\,\text{m} - 3\,\text{m}}{4\,\text{s} - 0\,\text{s}} = 0\,\text{m/s}$

  • Alternative discrete segment calculation:       $v = \frac{4\,\text{m} - 3\,\text{m}}{2\,\text{s} - 0\,\text{s}} = 0.5\,\text{m/s}$

  • Interval $b$ ($t = 4\,\text{s}$ to $t = 7\,\text{s}$):

  • Initial position at $t = 4\,\text{s}$: $x_i = 3\,\text{m}$

  • Final position at $t = 7\,\text{s}$: $x_f = 4\,\text{m}$

  • Average velocity calculation:       $v_{avg,b} = \frac{4\,\text{m} - 3\,\text{m}}{7\,\text{s} - 4\,\text{s}} = \frac{1\,\text{m}}{3\,\text{s}} \approx 0.33\,\text{m/s}$

  • Alternative discrete segment calculation:       $v = \frac{-3\,\text{m} - 4\,\text{m}}{6\,\text{s} - 2\,\text{s}} = \frac{-7\,\text{m}}{4\,\text{s}} = -1.75\,\text{m/s}$

  • Interval $c$ ($t = 7\,\text{s}$ to $t = 10\,\text{s}$):

  • Initial position at $t = 7\,\text{s}$: $x_i = 4\,\text{m}$

  • Final position at $t = 10\,\text{s}$: $x_f = -2\,\text{m}$

  • Average velocity calculation:       $v_{avg,c} = \frac{-2\,\text{m} - 4\,\text{m}}{10\,\text{s} - 7\,\text{s}} = \frac{-6\,\text{m}}{3\,\text{s}} = -2\,\text{m/s}$

  • Alternative discrete segment calculation:       $v = \frac{1\,\text{m} - (-3\,\text{m})}{10\,\text{s} - 6\,\text{s}} = \frac{4\,\text{m}}{4\,\text{s}} = 1\,\text{m/s}$

  • Interval $d$ ($t = 10\,\text{s}$ to $t = 13\,\text{s}$):

  • Initial position at $t = 10\,\text{s}$: $x_i = -2\,\text{m}$

  • Final position at $t = 13\,\text{s}$: $x_f = 1\,\text{m}$

  • Average velocity calculation:       $v_{avg,d} = \frac{1\,\text{m} - (-2\,\text{m})}{13\,\text{s} - 10\,\text{s}} = \frac{3\,\text{m}}{3\,\text{s}} = 1\,\text{m/s}$

  • Alternative discrete segment calculation:       $v = \frac{1\,\text{m} - 1\,\text{m}}{14\,\text{s} - 10\,\text{s}} = 0\,\text{m/s}$

Calculation of Average Acceleration

• Problem Statement:

  • An object moves along the $x$ axis. At time $t = 5.0\,\text{s}$, its position is $x = +3.0\,\text{m}$ with velocity $+5.0\,\text{m/s}$. At time $t = 8.0\,\text{s}$, its position is $x = +9.0\,\text{m}$ with velocity $-1.0\,\text{m/s}$. Find its average acceleration during the time interval $5.0\,\text{s} < t < 8.0\,\text{s}$.

• Given Quantities:

  • Initial time: $t_1 = 5.0\,\text{s}$

  • Initial position: $x_1 = +3.0\,\text{m}$

  • Initial velocity: $v_1 = +5.0\,\text{m/s}$

  • Final time: $t_2 = 8.0\,\text{s}$

  • Final position: $x_2 = +9.0\,\text{m}$

  • Final velocity: $v_2 = -1.0\,\text{m/s}$

• Average Acceleration Formula:

  • Average acceleration measures the rate of change of velocity over a given time interval:     $a_{avg} = \frac{v_2 - v_1}{t_2 - t_1}$

• Step-by-Step Solution:

  • Substituting values into the acceleration formula:     $a_{avg} = \frac{-1.0\,\text{m/s} - 5.0\,\text{m/s}}{8.0\,\text{s} - 5.0\,\text{s}}$     $a_{avg} = \frac{-6.0\,\text{m/s}}{3.0\,\text{s}}$     $a_{avg} = -2.0\,\text{m/s}^2$

Vertical Free-Fall Kinematics

• Problem Statement:

  • A ball is launched directly upward from ground level with an initial speed of $20\,\text{m/s}$. Air resistance is negligible.

  • $(a)$ How long is the ball in the air?

  • $(b)$ What is the greatest height reached by the ball?

  • $(c)$ How many seconds after launch is the ball $15\,\text{m}$ above the release point?

• Initial Conditions and Parameters:

  • Initial position: $x_0 = 0\,\text{m}$

  • Initial velocity: $v_0 = 20\,\text{m/s}$

  • Acceleration due to gravity: $a = -9.8\,\text{m/s}^2$

• Part (a) Solution - Total Time in Air:

  • Kinematic velocity equation:     $v = v_0 + a t$

  • At the highest point of travel, the instantaneous velocity is zero ($v = 0\,\text{m/s}$):     $0 = 20\,\text{m/s} + (-9.8\,\text{m/s}^2) t_{up}$     $9.8 t_{up} = 20$     $t_{up} = \frac{20}{9.8} \approx 2.04\,\text{s}$

  • Total flight time equals twice the time required to reach the apex:     $t_{total} = 2 \times t_{up} = 2 \times 2.04\,\text{s} = 4.08\,\text{s}$

• Part (b) Solution - Maximum Height Reached:

  • Kinematic relation between velocity, acceleration, and displacement:     $v^2 = v_0^2 + 2 a (x - x_0)$

  • Setting final velocity $v = 0\,\text{m/s}$ at peak height:     $0^2 = (20\,\text{m/s})^2 + 2 (-9.8\,\text{m/s}^2) x_{max}$     $0 = 400 + 2(-9.8) x_{max}$     $19.6 x_{max} = 400$     $x_{max} = \frac{400}{19.6} \approx 20.4\,\text{m}$

• Part (c) Solution - Elapsed Time at Height $15\,\text{m}$:

  • Kinematic position-time equation:     $x = x_0 + v_0 t + \frac{1}{2} a t^2$

  • Substituting $x = 15\,\text{m}$, $v_0 = 20\,\text{m/s}$, and $a = -9.8\,\text{m/s}^2$:     $15 = 20 t - 4.9 t^2$     $4.9 t^2 - 20 t + 15 = 0$

  • Solving the quadratic equation $a t^2 + b t + c = 0$ with $a = 4.9$, $b = -20$, $c = 15$:     $t = \frac{-b \pm \sqrt{b^2 - 4 a c}}{2 a}$     $t = \frac{20 \pm \sqrt{(-20)^2 - 4(4.9)(15)}}{2(4.9)}$     $t = \frac{20 \pm \sqrt{400 - 294}}{9.8}$     $t = \frac{20 \pm \sqrt{106}}{9.8}$

  • Evaluating both real roots:

  • First root (ascending path): $t_1 = \frac{20 - \sqrt{106}}{9.8} \approx 1.02\,\text{s}$

  • Second root (descending path): $t_2 = \frac{20 + \sqrt{106}}{9.8} \approx 3.00\,\text{s}$

Reaction Time and Hard Braking Kinematics

• Problem Statement:

  • An automobile under hard braking loses speed at a rate of $7.0\,\text{m/s}^2$. Reaction time to engage brakes is $0.50\,\text{s}$. A school zone speed limit requires all cars to be able to stop in $4.0\,\text{m}$.

  • $(a)$ What maximum initial speed does this imply for an automobile in this zone?

  • $(b)$ What fraction of the $4.0\,\text{m}$ distance is due to reaction time?

• Given Parameters:

  • Deceleration rate during braking: $a = -7.0\,\text{m/s}^2$

  • Reaction time before applying brakes: $t_r = 0.50\,\text{s}$

  • Maximum stopping distance: $d_{total} = 4.0\,\text{m}$

• Part (a) Solution - Maximum Allowed Initial Speed $v_0$:

  • Reaction Distance ($d_r$):

  • Distance traveled at constant velocity $v_0$ prior to brake application:       $d_r = v_0 \times t_r = 0.50 v_0$

  • Braking Distance ($d_b$):

  • Using kinematic equation $v^2 = v_0^2 + 2 a d_b$ with final velocity $v = 0\,\text{m/s}$:       $0 = v_0^2 + 2(-7.0) d_b$       $14 d_b = v_0^2$       $d_b = \frac{v_0^2}{14}$

  • Combined Stopping Distance Quadratic Equation:     $d_{total} = d_r + d_b$     $0.50 v_0 + \frac{v_0^2}{14} = 4.0$

  • Multiplying through by $14$ to convert to standard quadratic form:     $v_0^2 + 7 v_0 - 56 = 0$

  • Applying quadratic formula:     $v_0 = \frac{-7 + \sqrt{7^2 - 4(1)(-56)}}{2(1)}$     $v_0 = \frac{-7 + \sqrt{49 + 224}}{2}$     $v_0 = \frac{-7 + \sqrt{273}}{2}$     $v_0 \approx 4.76\,\text{m/s}$

• Part (b) Solution - Fraction of Distance Due to Reaction Time:

  • Calculating distance traveled during reaction period:     $d_r = v_0 \times t_r = 4.76\,\text{m/s} \times 0.50\,\text{s} = 2.38\,\text{m}$

  • Computing fraction relative to total stopping distance of $4.0\,\text{m}$:     $\text{Fraction} = \frac{d_r}{d_{total}} = \frac{2.38\,\text{m}}{4.0\,\text{m}} = 0.595$

  • Expressed as a percentage:     $\text{Percentage} = 59.5\%$