1. Fundamental Concepts of Motion
Motion along a straight line (one-dimensional kinematics) describes the position, velocity, and acceleration of a particle as a function of time. We use a single spatial coordinate, $x(t)$, to define the position of the particle relative to a chosen origin.
1.1. Position and Displacement
The position $x$ is a coordinate vector, giving the location relative to the origin. The SI unit is meters ($\text{m}$). The displacement ($\Delta x$) is the change in position over a time interval $\Delta t = t_f - t_i$.
Displacement is a vector quantity (sign matters), while the total distance traveled is a scalar quantity, often greater than or equal to the magnitude of the displacement, $|\Delta x|$.
1.2. Average Velocity and Speed
The average velocity ($\bar{v}$) is the ratio of displacement to the time interval:
The average velocity is a vector quantity. The average speed is the ratio of the total distance traveled to the total time interval, a scalar quantity.
2. Instantaneous Velocity and Acceleration
2.1. Instantaneous Velocity ($v$)
Instantaneous velocity is the limit of the average velocity as the time interval approaches zero. It is the first time derivative of the position function $x(t)$.
The magnitude of the instantaneous velocity, $|v(t)|$, is the instantaneous speed. The velocity function is the slope of the position-time graph.
2.2. Acceleration ($a$)
Instantaneous acceleration ($a$) is the rate of change of instantaneous velocity. It is the first derivative of the velocity function, and the second derivative of the position function $x(t)$.
Acceleration is the slope of the velocity-time graph. Its SI unit is meters per second squared ($\text{m}/\text{s}^2$).
2.3. Advanced Relationship (Chain Rule)
Acceleration can also be expressed in terms of position by using the chain rule, a crucial form for non-time-dependent problems:
3. Constant Acceleration and Kinematic Equations
When acceleration is constant ($a(t) = a = \text{constant}$), the motion equations can be solved analytically using integration. We assume motion starts at $t=0$ with initial velocity $v_0$ and initial position $x_0$.
3.1. Velocity as a Function of Time ($v(t)$)
Starting with the definition of acceleration, $a = dv/dt$, we separate variables and integrate over the boundaries:
Integration Proof for $v(t)$:
$$\frac{dv}{dt} = a \implies dv = a dt$$ $$\int_{v_0}^{v} dv' = \int_{0}^{t} a dt'$$ $$[v']_{v_0}^{v} = a [t']_{0}^{t}$$ $$v - v_0 = a t$$ $$v = v_0 + at \quad \text{(3.1)}$$3.2. Position as a Function of Time ($x(t)$)
Using the definition of velocity, $v = dx/dt$, and substituting the derived expression for $v(t)$ (3.1):
Integration Proof for $x(t)$:
$$\frac{dx}{dt} = v(t) = v_0 + at \implies dx = (v_0 + at) dt$$ $$\int_{x_0}^{x} dx' = \int_{0}^{t} (v_0 + at') dt'$$ $$[x']_{x_0}^{x} = \int_{0}^{t} v_0 dt' + \int_{0}^{t} a t' dt'$$ $$x - x_0 = v_0 [t']_{0}^{t} + a \left[\frac{(t')^2}{2}\right]_{0}^{t}$$ $$x - x_0 = v_0 t + \frac{1}{2} a t^2$$ $$x = x_0 + v_0 t + \frac{1}{2} a t^2 \quad \text{(3.2)}$$3.3. Velocity as a Function of Position ($v(x)$)
This third kinematic equation is derived by substituting time $t$ from (3.1) into (3.2), or more elegantly using the advanced calculus relation (2.3): $a = v(dv/dx)$.
Integration Proof for $v(x)$:
$$a = v \frac{dv}{dx} \implies a dx = v dv$$ $$\int_{x_0}^{x} a dx' = \int_{v_0}^{v} v' dv'$$ $$a [x']_{x_0}^{x} = \left[\frac{(v')^2}{2}\right]_{v_0}^{v}$$ $$a (x - x_0) = \frac{1}{2} v^2 - \frac{1}{2} v_0^2$$ $$v^2 = v_0^2 + 2a (x - x_0) \quad \text{(3.3)}$$4. Practice Problems
These problems require analytical application of the kinematic equations, calculus, and advanced conceptual understanding of one-dimensional motion.
4.1. Intermediate Problems (5)
- A particle's position is given by $x(t) = 2.0 t^3 - 5.0 t^2 + 7.0 \text{ m}$. Determine the instantaneous velocity and instantaneous acceleration at $t = 2.0 \text{ s}$.
- A train accelerates uniformly from rest at $a = 1.5 \text{ m/s}^2$ for $20 \text{ s}$. It then travels at a constant velocity for $60 \text{ s}$, and finally decelerates at a constant rate of $2.0 \text{ m/s}^2$ until it stops. Find the total distance traveled.
- Two cars, A and B, are initially $100 \text{ m}$ apart. Car A starts from rest with $a_A = 2.0 \text{ m/s}^2$ toward B. Car B travels towards A at a constant velocity of $15 \text{ m/s}$. Find the time $t$ at which they meet.
- A bullet traveling at $400 \text{ m/s}$ penetrates a wooden plank to a depth of $5.0 \text{ cm}$. Assuming constant acceleration (deceleration), find the magnitude of the acceleration.
- A sprinter can maintain a maximum speed of $12 \text{ m/s}$. If they accelerate uniformly from rest at $4.0 \text{ m/s}^2$, how long does it take for them to reach maximum speed, and what distance do they cover in that time?
4.2. Advanced Problems (3)
- The acceleration of a particle moving along the x-axis is given by $a(t) = k t$, where $k$ is a constant. If the particle starts from rest at the origin ($x_0=0, v_0=0$), derive the position function $x(t)$.
- A rocket is launched vertically with an acceleration of $a(t) = c / (t+1)$, where $c$ is a constant. If $v_0 = 0$, find the total distance $x$ traveled during the time interval $[0, T]$.
- A particle moving along the x-axis has a velocity dependent acceleration $a(v) = -k v^2$, where $k > 0$ is a constant. If $v(0) = v_0$, find the time $t$ required for the velocity to decrease to $v_0/2$.
4.3. Irodov-like Problems (2)
- Two particles, 1 and 2, move with constant accelerations $a_1$ and $a_2$ (both non-zero). They start at the same position and time. After time $T$, their velocities are $v_1$ and $v_2$. Find the acceleration of particle 1, $a_1$, in terms of $v_1$, $v_2$, and $T$, given that their average velocities over the interval $[0, T]$ are the same.
- A particle starts from rest and moves with acceleration $a = 3 \sqrt{v}$, where $v$ is the instantaneous speed. Find the distance traveled by the particle when it reaches a speed of $v_f = 9 \text{ m/s}$.
5. Solutions to Practice Problems
The solutions below apply calculus and the fundamental kinematic definitions rigorously to solve each problem.
5.1. Intermediate Solutions
Problem 1: Time Derivatives
Position: $x(t) = 2.0 t^3 - 5.0 t^2 + 7.0$
Velocity: $v(t) = \frac{dx}{dt} = \frac{d}{dt} (2.0 t^3 - 5.0 t^2 + 7.0) = 6.0 t^2 - 10.0 t$
Acceleration: $a(t) = \frac{dv}{dt} = \frac{d}{dt} (6.0 t^2 - 10.0 t) = 12.0 t - 10.0$
At $t = 2.0 \text{ s}$:
$$v(2.0) = 6.0 (2.0)^2 - 10.0 (2.0) = 24.0 - 20.0 = 4.0 \text{ m/s}$$ $$a(2.0) = 12.0 (2.0) - 10.0 = 24.0 - 10.0 = 14.0 \text{ m/s}^2$$Problem 2: Multi-Stage Motion
Phase 1 (Acceleration): $v_0 = 0, a_1 = 1.5 \text{ m/s}^2, t_1 = 20 \text{ s}$.
$$v_1 = v_0 + a_1 t_1 = 0 + (1.5)(20) = 30.0 \text{ m/s}$$ $$x_1 = v_0 t_1 + \frac{1}{2} a_1 t_1^2 = 0 + \frac{1}{2} (1.5) (20)^2 = 300.0 \text{ m}$$Phase 2 (Constant Velocity): $v_2 = 30.0 \text{ m/s}, t_2 = 60 \text{ s}$.
$$x_2 = v_2 t_2 = (30.0) (60) = 1800.0 \text{ m}$$Phase 3 (Deceleration): $v_i = 30.0 \text{ m/s}, v_f = 0, a_3 = -2.0 \text{ m/s}^2$. Using $v^2 = v_0^2 + 2a \Delta x$:
$$0^2 = (30.0)^2 + 2 (-2.0) x_3 \implies 4.0 x_3 = 900.0 \implies x_3 = 225.0 \text{ m}$$Total Distance: $x_{total} = x_1 + x_2 + x_3 = 300.0 + 1800.0 + 225.0 = 2325.0 \text{ m}$
Problem 3: Meeting Time
Let $x=0$ be the starting point of Car A. The meeting occurs when $x_A = x_B$.
Car A: $x_A(t) = x_{A0} + v_{A0} t + \frac{1}{2} a_A t^2 = 0 + 0 + \frac{1}{2} (2.0) t^2 = t^2$
Car B: $x_B(t) = x_{B0} + v_{B0} t = 100 + (-15) t = 100 - 15 t$
Set $x_A = x_B$:
$$t^2 = 100 - 15 t \implies t^2 + 15 t - 100 = 0$$Using the quadratic formula, $t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:
$$t = \frac{-15 \pm \sqrt{15^2 - 4(1)(-100)}}{2} = \frac{-15 \pm \sqrt{625}}{2} = \frac{-15 \pm 25}{2}$$The positive solution is $t = \frac{10}{2} = 5.0 \text{ s}$
Problem 4: Deceleration in a Medium
Given: $v_0 = 400 \text{ m/s}, v_f = 0, \Delta x = 0.05 \text{ m}$. Use $v^2 = v_0^2 + 2a \Delta x$:
$$0^2 = (400)^2 + 2 a (0.05)$$ $$0 = 160,000 + 0.10 a$$ $$a = \frac{-160,000}{0.10} = -1,600,000 \text{ m/s}^2$$The magnitude of acceleration is $|\vec{a}| = 1.6 \times 10^6 \text{ m/s}^2$
Problem 5: Time and Distance to Max Speed
Given: $v_0 = 0, v_f = 12 \text{ m/s}, a = 4.0 \text{ m/s}^2$.
Time to reach maximum speed (using $v = v_0 + at$):
$$12 = 0 + (4.0) t \implies t = \frac{12}{4.0} = 3.0 \text{ s}$$Distance covered (using $v^2 = v_0^2 + 2a \Delta x$):
$$(12)^2 = 0^2 + 2 (4.0) \Delta x$$ $$144 = 8.0 \Delta x \implies \Delta x = \frac{144}{8.0} = 18.0 \text{ m}$$5.2. Advanced Solutions
Problem 6: Position with Time-Dependent Acceleration
Step 1: Find $v(t)$ by integrating $a(t) = kt$ with $v_0 = 0$:
$$v(t) = \int a(t) dt = \int kt dt = \frac{1}{2} k t^2 + C_1$$Since $v(0)=0$, $C_1=0$, so $v(t) = \frac{1}{2} k t^2$.
Step 2: Find $x(t)$ by integrating $v(t)$ with $x_0 = 0$:
$$x(t) = \int v(t) dt = \int \frac{1}{2} k t^2 dt = \frac{1}{2} k \left(\frac{t^3}{3}\right) + C_2$$Since $x(0)=0$, $C_2=0$.
$$x(t) = \frac{1}{6} k t^3$$Problem 7: Distance with Logarithmic Acceleration
Step 1: Find $v(t)$ by integrating $a(t) = c/(t+1)$ with $v_0 = 0$:
$$v(t) = \int_{0}^{t} \frac{c}{t'+1} dt' = c [\ln|t'+1|]_{0}^{t}$$ $$v(t) = c (\ln(t+1) - \ln(1)) = c \ln(t+1)$$Step 2: Find the distance $x$ by integrating $v(t)$ from $0$ to $T$:
$$x = \int_{0}^{T} v(t) dt = \int_{0}^{T} c \ln(t+1) dt$$Use substitution: $u = t+1$, $du = dt$. Limits: $t=0 \to u=1$, $t=T \to u=T+1$.
$$x = c \int_{1}^{T+1} \ln(u) du$$Applying the standard integral $\int \ln(u) du = u \ln(u) - u$:
$$x = c [u \ln(u) - u]_{1}^{T+1}$$ $$x = c \left[ \left((T+1) \ln(T+1) - (T+1)\right) - \left(1 \ln(1) - 1\right) \right]$$ $$x = c \left[ (T+1) \ln(T+1) - T - 1 + 1 \right]$$ $$x = c \left[ (T+1) \ln(T+1) - T \right]$$Problem 8: Time with Velocity-Dependent Acceleration
Use the definition $a = dv/dt$. Separate variables and integrate:
$$a(v) = -k v^2 \implies \frac{dv}{dt} = -k v^2$$ $$\int_{v_0}^{v_0/2} \frac{dv}{v^2} = \int_{0}^{t} -k dt$$ $$\left[-\frac{1}{v}\right]_{v_0}^{v_0/2} = -k [t]_{0}^{t}$$ $$-\left(\frac{1}{v_0/2} - \frac{1}{v_0}\right) = -k t$$ $$-\left(\frac{2}{v_0} - \frac{1}{v_0}\right) = -k t$$ $$-\frac{1}{v_0} = -k t$$ $$t = \frac{1}{k v_0}$$5.3. Irodov-like Solutions
Problem 9: Average Velocity and Acceleration Relation
For constant acceleration, the average velocity is $\bar{v} = (v_0 + v_f)/2$. Since $v_{10} = 0$ and $v_{20} = 0$, the average velocities are:
$$\bar{v}_1 = \frac{v_1}{2} \quad \text{and} \quad \bar{v}_2 = \frac{v_2}{2}$$The problem states $\bar{v}_1 = \bar{v}_2$, which means $v_1 = v_2$.
Using the kinematic equation $v = v_0 + at$, we have:
$$v_1 = a_1 T \quad \text{and} \quad v_2 = a_2 T$$Since $v_1 = v_2$:
$$a_1 T = a_2 T \implies a_1 = a_2$$The average velocity is also defined as $\bar{v} = \Delta x / \Delta t$. For constant acceleration from rest, $\Delta x = \frac{1}{2} a T^2$.
$$\bar{v}_1 = \frac{\frac{1}{2} a_1 T^2}{T} = \frac{1}{2} a_1 T$$Equating this to the first result: $\frac{1}{2} a_1 T = \frac{v_1}{2} \implies a_1 = \frac{v_1}{T}$.
The condition $v_1=v_2$ implies that $a_1$ is also equal to $v_2/T$.
$$a_1 = \frac{v_1}{T}$$Problem 10: Distance with Velocity-Dependent Acceleration
Use the advanced form $a = v(dv/dx)$. Separate variables and integrate:
$$a = 3 \sqrt{v} \implies v \frac{dv}{dx} = 3 \sqrt{v}$$ $$\frac{v}{\sqrt{v}} dv = 3 dx \implies \sqrt{v} dv = 3 dx$$Integrate from $x_0=0$ to $x$ and $v_0=0$ to $v_f=9$:
$$\int_{0}^{9} v^{1/2} dv = \int_{0}^{x} 3 dx$$ $$\left[\frac{v^{3/2}}{3/2}\right]_{0}^{9} = 3 [x]_{0}^{x}$$ $$\frac{2}{3} (9^{3/2}) = 3 x$$ $$\frac{2}{3} (27) = 3 x$$ $$18 = 3 x$$ $$x = 6.0 \text{ m}$$