Graphical Kinematics and Equations

Graphical Representation Of Motion

Graphical methods provide a visual way to understand and analyze the motion of objects. By plotting quantities like position, velocity, and time against each other, we can extract important information about the motion, such as speed, acceleration, and displacement.

Distance–Time Graphs

A distance-time graph plots the distance travelled by an object against time. The time is typically on the horizontal axis (x-axis), and the distance is on the vertical axis (y-axis).

• Slope of a distance-time graph: The slope of a distance-time graph represents the speed of the object.

  $ \text{Speed} = \frac{\text{Change in Distance}}{\text{Change in Time}} = \text{Slope} $

• Interpretation of different slopes:
  • Horizontal line: The distance is not changing with time. This means the object is at rest. The speed is zero.
  • Straight line with a positive slope: The distance is increasing linearly with time. This indicates uniform motion with a constant positive speed. The steeper the slope, the greater the speed.
  • Straight line with a negative slope: This would imply distance decreasing with time, which is not physically possible for distance (as it's always positive path length). If the graph were position-time, a negative slope would indicate motion in the negative direction.
  • Curved line: The slope is changing with time, indicating that the speed is not constant. This represents non-uniform motion (accelerated motion).

Example: If a graph shows a line starting from 0 km at 0 seconds, going to 10 km at 10 seconds, and then remaining at 10 km until 20 seconds, it means the object moved at a constant speed for the first 10 seconds and then stopped.

Velocity-Time Graphs

A velocity-time graph plots the velocity of an object against time. Time is on the horizontal axis, and velocity is on the vertical axis.

• Slope of a velocity-time graph: The slope of a velocity-time graph represents the acceleration of the object.

  $ \text{Acceleration} = \frac{\text{Change in Velocity}}{\text{Change in Time}} = \text{Slope} $

• Area under a velocity-time graph: The area under a velocity-time graph represents the displacement (or distance travelled if velocity is always positive) of the object.

  $ \text{Displacement} = \text{Area under the curve} $

• Interpretation of different slopes and areas:
  • Horizontal line: Velocity is constant. This means the acceleration is zero (uniform motion). The area under the line is a rectangle, representing $ \text{velocity} \times \text{time} = \text{displacement} $.
  • Straight line with a positive slope: Velocity is increasing linearly with time. This indicates uniform acceleration in the direction of motion. The area under the line is a trapezoid (or triangle + rectangle), representing the displacement.
  • Straight line with a negative slope: Velocity is decreasing linearly with time. This indicates uniform deceleration (acceleration in the opposite direction of motion). The area might be positive or negative depending on whether the velocity is positive or negative.
  • Curved line: The slope is changing, indicating non-uniform acceleration. The area under the curve still represents displacement, but it might require calculus to calculate precisely.

Example: A velocity-time graph showing a straight line sloping upwards from (0,0) to (10s, 20m/s) means the object started from rest and had a constant acceleration of $ (20-0)/(10-0) = 2 \, \text{m/s}^2 $. The displacement during this time would be the area of the triangle: $ \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 10 \, \text{s} \times 20 \, \text{m/s} = 100 \, \text{m} $.

Equations Of Motion By Graphical Method

The standard equations of motion for uniformly accelerated linear motion can be derived using velocity-time graphs.

Equation For Velocity-time Relation ($ v = u + at $)

Consider a velocity-time graph for an object starting with initial velocity $ u $ at time $ t=0 $ and reaching a final velocity $ v $ at time $ t $. Assume uniform acceleration $ a $.

The graph is a straight line. The initial velocity $ u $ is the intercept on the y-axis. The final velocity $ v $ is reached at time $ t $.

The acceleration $ a $ is the slope of this line:

$ a = \text{Slope} = \frac{\text{Change in velocity}}{\text{Change in time}} = \frac{v - u}{t - 0} $

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

Rearranging this equation gives the velocity-time relation:

$ at = v - u $

$ \mathbf{v = u + at} $

Equation For Position-time Relation ($ s = ut + \frac{1}{2}at^2 $)

The displacement (or distance $ s $) covered by the object during time $ t $ is represented by the area under the velocity-time graph between $ t=0 $ and time $ t $.

The area under the graph is a trapezoid (or a rectangle plus a triangle).

$ \text{Area} = \text{Area of rectangle OADC} + \text{Area of triangle ABD} $

Where O is the origin, A is at (0, u), B is at (t, v), C is at (t, 0), and D is at (0, v).

Area of rectangle OADC = base × height = $ t \times u $.

Area of triangle ABD = $ \frac{1}{2} \times \text{base} \times \text{height} $. The base is $ t $, and the height is $ (v-u) $. From the first equation, we know $ v-u = at $. So, height is $ at $.

Area of triangle ABD = $ \frac{1}{2} \times t \times (at) = \frac{1}{2}at^2 $.

Therefore, the total displacement $ s $ is:

$ s = \text{Area of rectangle} + \text{Area of triangle} $

$ s = ut + \frac{1}{2}at^2 $

This gives us the position-time relation.

Equation For Position–velocity Relation ($ v^2 = u^2 + 2as $)

We can also find the area under the velocity-time graph as a single trapezoid.

$ s = \text{Area of trapezoid OABC} $

The area of a trapezoid is given by $ \frac{1}{2} \times (\text{sum of parallel sides}) \times \text{height} $.

Here, the parallel sides are the initial velocity $ u $ and the final velocity $ v $. The height is the time interval $ t $.

$ s = \frac{1}{2} (u + v) t $

Now, we need to eliminate $ t $ from this equation. From the first equation ($ v = u + at $), we have $ t = \frac{v - u}{a} $.

Substitute this expression for $ t $ into the area equation:

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

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

$ s = \frac{1}{2} \frac{v^2 - u^2}{a} $

Rearranging this equation gives the position-velocity relation:

$ 2as = v^2 - u^2 $

$ \mathbf{v^2 = u^2 + 2as} $

These three equations are the fundamental kinematic equations for uniformly accelerated motion.