derive, from the definitions of velocity and acceleration, equations that represent uniformly accelerated motion in a straight line

1️⃣ Definitions

Velocity is the rate of change of position:

\$v(t)=\frac{dx}{dt}\$

Acceleration is the rate of change of velocity:

\$a(t)=\frac{dv}{dt}\$

For uniformly accelerated motion, a is constant (doesn’t depend on time).

2️⃣ Deriving the First Equation: 𝑣 = 𝑣₀ + a𝑡

1. Start with the definition of acceleration: \$a=\frac{dv}{dt}\$
2. Because a is constant, integrate both sides with respect to time:
3. \$\int a\,dt = \int dv\$
4. \$a\,t + C_1 = v\$
5. At time t = 0, velocity is the initial velocity \(v0\). So \(C1 = v_0\).
6. Thus, \$v = v_0 + a\,t\$

📐 Analogy: Think of a car that starts from rest and speeds up at a constant rate. After 5 s, its speed is just the starting speed plus 5 s times the acceleration.

3️⃣ Deriving the Second Equation: 𝑥 = 𝑥₀ + 𝑣₀𝑡 + ½𝑎𝑡²

1. We know \(v = \frac{dx}{dt}\). Substitute the expression for \(v\) from the first equation:
2. \$\frac{dx}{dt} = v_0 + a\,t\$
3. Integrate with respect to time:
4. \$\int dx = \int (v_0 + a\,t)\,dt\$
5. \$x + C2 = v0\,t + \frac{1}{2}a\,t^2\$
6. At t = 0, position is \(x0\). Hence \(C2 = x_0\).
7. Therefore, \$x = x0 + v0\,t + \frac{1}{2}a\,t^2\$

🚗 Example: A car starts from \(x0 = 0\) m with \(v0 = 0\) m/s and accelerates at \(a = 3\) m/s². After 4 s, its position is \(x = 0 + 0·4 + ½·3·4² = 24\) m.

4️⃣ Deriving the Third Equation: 𝑣² = 𝑣₀² + 2𝑎(𝑥−𝑥₀)

1. Start from \(a = \frac{dv}{dt}\) and use \(v = \frac{dx}{dt}\) to eliminate time:
2. \$a = \frac{dv}{dt} = \frac{dv}{dx}\frac{dx}{dt} = v\,\frac{dv}{dx}\$
3. Rearrange and integrate:
4. \$v\,dv = a\,dx\$
5. \$\int v\,dv = \int a\,dx\$
6. \$\frac{1}{2}v^2 + C_3 = a\,x\$
7. At \(x = x0\), velocity is \(v0\). Thus \(C3 = \frac{1}{2}v0^2 - a\,x_0\).
8. Combine terms to get \$v^2 = v0^2 + 2a(x - x0)\$

🧮 Use Case: When you know the distance travelled and want to find the final speed without time.

5️⃣ Quick Example: A Ball Dropped from a Height

Let a ball be dropped from rest at height \(h = 20\) m. Take upward as positive, so \(a = -9.8\) m/s².

• Using \(v^2 = v0^2 + 2a(x - x0)\) with \(v0 = 0\), \(x0 = 20\) m, \(x = 0\) m:
• \$v^2 = 0 + 2(-9.8)(0 - 20) = 392\$
• Thus \(v = \sqrt{392} \approx 19.8\) m/s downward.

📐 Check with time: Using \(v = v0 + a t\), \(t = \frac{v - v0}{a} = \frac{19.8}{9.8} \approx 2.02\) s, which matches the free‑fall time.

📚 Examination Tips

• Always state the assumption that acceleration is constant before applying any equation.
• Check units: m/s for velocity, m/s² for acceleration, m for distance.
• Remember the sign convention: choose a direction (upward, rightward) as positive.
• When solving for time, use the equation that does not involve the variable you’re solving for (e.g., use \(v = v_0 + a t\) to find \(t\) if \(v\) is known).
• In multi‑step problems, write down all known values and the equation you’ll use before plugging numbers.
• Practice converting between the three main equations; they’re all equivalent but useful in different contexts.