Kinematic Equations for Uniformly Accelerated Motion
Key Equations
For an object moving with constant acceleration, the following equations relate displacement, time, velocity, and acceleration:
- Final velocity: \( v = v_0 + at \)
- Displacement: \( x = x_0 + v_0 t + \frac{1}{2} a t^2 \)
- Velocity-displacement relation: \( v^2 = v_0^2 + 2a(x – x_0) \)
Here, \( v_0 \) is the initial velocity, \( v \) is the final velocity, \( a \) is acceleration, \( t \) is time, \( x_0 \) is the initial position, and \( x \) is the final position.
Graphical Interpretation
The area under a velocity-time (v-t) graph represents the displacement of the object:
- For uniform motion (constant velocity), the area is a rectangle: \( \text{Displacement} = u \times T \).
- For uniformly accelerated motion, the area combines a rectangle and a triangle: \( \text{Displacement} = v_0 t + \frac{1}{2} a t^2 \).
Free Fall Motion
When an object is in free fall (neglecting air resistance), its motion is governed by gravity (\( g = 9.8 \, \text{m/s}^2 \) downward). The equations simplify to:
- \( v = -gt \) (velocity as a function of time)
- \( y = -\frac{1}{2} g t^2 \) (displacement as a function of time)
- \( v^2 = -2g y \) (velocity as a function of displacement)
Note: The negative sign indicates the downward direction.
Stopping Distance and Reaction Time
Stopping distance (\( d_s \)) is the distance a vehicle travels before coming to rest after braking. It depends on initial velocity (\( v_0 \)) and deceleration (\( -a \)):
\[ d_s = \frac{-v_0^2}{2a} \]
Reaction time (\( t_r \)) is the time taken to respond to a situation. For a ruler dropped in free fall, the reaction time can be calculated using:
\[ t_r = \sqrt{\frac{2d}{g}} \]
where \( d \) is the distance the ruler falls.
Important NEET Concepts
- Equations of motion: Memorize \( v = v_0 + at \), \( x = v_0 t + \frac{1}{2} a t^2 \), and \( v^2 = v_0^2 + 2ax \). These are frequently tested.
- Free fall: Understand how to apply the kinematic equations with \( a = -g \).
- Graphical analysis: Interpret v-t graphs to find displacement and acceleration.
- Stopping distance: Know how initial velocity and deceleration affect the distance required to stop.
- Reaction time: Be familiar with the simple experiment to measure reaction time using free fall.
Example Problems
Example 1: A ball is thrown upward with \( v_0 = 20 \, \text{m/s} \). How high does it rise?
Solution: Use \( v^2 = v_0^2 + 2a(y – y_0) \) with \( v = 0 \) and \( a = -g \). Solve for \( y \).
Example 2: A car decelerates at \( -5 \, \text{m/s}^2 \). If \( v_0 = 20 \, \text{m/s} \), what is the stopping distance?
Solution: Use \( d_s = \frac{-v_0^2}{2a} \). Plug in the values to find \( d_s \).
Summary
- Motion with constant acceleration follows predictable patterns described by kinematic equations.
- Graphs (v-t, x-t) help visualize motion and solve problems.
- Free fall and braking distances are practical applications of these concepts.
