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

Equations of Motion

In this section we derive the kinematic equations that describe uniformly accelerated motion in a straight line, starting from the fundamental definitions of velocity and acceleration.

1. Definitions

• Displacement \$s\$ (m) – the change in position along a straight line.
• Velocity \$v\$ (m s\(^{-1}\)) – the rate of change of displacement:

  \$v = \frac{ds}{dt}\$

• Acceleration \$a\$ (m s\(^{-2}\)) – the rate of change of velocity:

  \$a = \frac{dv}{dt}\$

• Initial velocity \$u\$ – velocity at \$t = 0\$.
• Final velocity \$v\$ – velocity at a later time \$t\$.

2. Constant (Uniform) Acceleration

For uniformly accelerated motion the acceleration \$a\$ is constant. This allows us to integrate the definitions directly.

2.1 Derivation of \$v = u + at\$

Starting from \$a = dv/dt\$ and treating \$a\$ as a constant:

1. Separate variables: \$dv = a\,dt\$.
2. Integrate from \$t = 0\$ (where \$v = u\$) to a general time \$t\$ (where \$v = v\$):

   \$\int{u}^{v} dv = a\int{0}^{t} dt\$

3. Perform the integrations:

   \$v - u = a t\$

4. Re‑arrange:

   \$\boxed{v = u + a t}\$

2.2 Derivation of \$s = ut + \tfrac{1}{2} a t^{2}\$

Use the definition \$v = ds/dt\$ and substitute the expression for \$v\$ obtained above:

1. Write \$ds/dt = u + a t\$.
2. Separate variables: \$ds = (u + a t)\,dt\$.
3. Integrate from \$t = 0\$ (where \$s = 0\$) to \$t\$ (where \$s = s\$):

   \$\int{0}^{s} ds = \int{0}^{t} (u + a t)\,dt\$

4. Perform the right‑hand integration:

   \$s = u t + \frac{1}{2} a t^{2}\$

5. Thus:

   \$\boxed{s = u t + \tfrac{1}{2} a t^{2}}\$

2.3 Derivation of \$v^{2} = u^{2} + 2 a s\$

Eliminate time \$t\$ between the two equations derived above.

1. From \$v = u + a t\$, solve for \$t\$:

   \$t = \frac{v - u}{a}\$

2. Substitute this \$t\$ into \$s = u t + \tfrac{1}{2} a t^{2}\$:

   \$s = u\left(\frac{v - u}{a}\right) + \frac{1}{2}a\left(\frac{v - u}{a}\right)^{2}\$

3. Simplify:

   \$s = \frac{u(v - u)}{a} + \frac{(v - u)^{2}}{2a}\$

4. Combine terms over a common denominator \$2a\$:

   \$2 a s = 2u(v - u) + (v - u)^{2}\$

5. Expand and rearrange:

   \$2 a s = v^{2} - u^{2}\$

6. Finally:

   \$\boxed{v^{2} = u^{2} + 2 a s}\$

2.4 Average‑velocity form \$s = \frac{u+v}{2}\,t\$

When acceleration is constant, the velocity changes linearly with time, so the average velocity \$\bar v\$ over the interval \$t\$ is the arithmetic mean of the initial and final velocities:

\$\bar v = \frac{u + v}{2}\$

Multiplying by the time interval gives the displacement:

\$\boxed{s = \frac{u+v}{2}\,t}\$

3. Summary of Kinematic Equations

4. Common Pitfalls

• Mixing up displacement \$s\$ with distance travelled – the equations apply to straight‑line displacement, not curved paths.
• Using the wrong sign for acceleration when the motion is opposite to the chosen positive direction.
• Assuming the equations hold for non‑constant acceleration; they are derived under the explicit condition \$a = \text{constant}\$.

5. Suggested Diagram

6. Quick Check Questions

1. If a car starts from rest and accelerates uniformly at \$3.0\ \text{m s}^{-2}\$ for \$5.0\ \text{s}\$, what is its final speed?
2. A stone is thrown vertically upward with an initial speed of \$20\ \text{m s}^{-1}\$. Using \$a = -9.8\ \text{m s}^{-2}\$, calculate the maximum height reached.
3. Given \$u = 4\ \text{m s}^{-1}\$, \$v = 20\ \text{m s}^{-1}\$ and \$a = 4\ \text{m s}^{-2}\$, find the displacement \$s\$ without using time.

These notes provide the logical chain from the basic definitions to the four standard equations of uniformly accelerated motion, ready for A‑Level study and examination practice.