Lecture 7 1d Kinematics

Area and Integration

Definition of A(t)

Let g(t) = df/dt.
A(t) denotes the area under the graph of g(t) from t = 0 to t.
By definition, A(t = 0) = 0.

Derivation of dA/dt

Using the definition of the area change:
- Change in area: ( \Delta A = A(t + \Delta t) - A(t) )
- Approximating ( \Delta A ): ( \Delta A \approx g(t) \Delta t )
- Thus, for small changes: ( \frac{\Delta A}{\Delta t} \approx g(t) + \text{small corrections} )
Taking the limit as ( \Delta t \rightarrow 0 ):
- Small corrections vanish, leading to ( \frac{dA}{dt} = g(t) )

Integration of g(t)

Integrating to find area:
- ( [A(t)]{0}^{t} = \int{0}^{t} g(t) dt )
- Since A(0) = 0, it simplifies to:
  - ( A(t) = \int_{0}^{t} g(t) dt )
General case for area under the curve from a to b:
- ( A = \int_{a}^{b} g(t) dt )

Examples

Example 1: Constant Function

For g(t) = H, 0 ≤ t ≤ W:
- Area: ( \int_{0}^{W} H dt = [Ht]_{0}^{W} = HW )

Example 2: Linear Function

For g(t) = t, 0 ≤ t ≤ 1:
- Area: ( \int_{0}^{1} t dt = \frac{1}{2} t^2 \bigg|_{0}^{1} = \frac{1}{2} )
- Consistent with the formula for the area of a triangle: ( \text{Area} = \frac{1}{2} \text{base} \times \text{height} )

Velocity/Time Diagrams

Analysis of Velocity/Time Diagram

Consider a general velocity/time diagram:
- Slope of curve represents acceleration: ( \frac{dv}{dt} = a )
- Relationship between position and time:
  - ( \frac{ds}{dt} = v(t) )
  - Position function: ( s(t) = \int_{0}^{t} v(t) dt = \text{area under the curve} )
Conclusion: Velocity-time diagrams effectively illustrate properties of one-dimensional motion.

Constant Acceleration

Area Calculation

With constant acceleration a:
- Area = sum of areas of a rectangle and triangle:
  - ( s = ut + \frac{1}{2}(v-u)t )
  - Rearranged: ( s = \frac{1}{2}(u+v)t )
Recalling that ( v = u + at ):
- Substituting in yields: ( s = ut + \frac{1}{2}at^2 )
- Confirms the standard formula for displacement in uniformly accelerated motion.