Multi-Stage Kinematics & Free-Fall

1. The Archetypal Case: Vertical Free-Fall

The motion of an object near the Earth's surface, under the sole influence of gravity (neglecting air resistance), is the most common example of motion with (nearly) constant acceleration.

This acceleration, known as the acceleration due to gravity, is denoted by $g$. Its direction is always vertically downwards.

• Standard value: $g\approx 9.8{\text{m/s}}^2$
• For JEE and many physics problems, it's often approximated as $g\approx 10{\text{m/s}}^2$ to simplify calculations. Always check if the problem specifies a value.

Sign Convention: The Most Critical Step

Since motion is in one dimension (vertical), we must define a coordinate system. A choice must be made for the positive direction. This choice is arbitrary, but once made, it must be applied consistently to all vector quantities: position ($y$), displacement ($\Delta y$), velocity ($v$), and acceleration ($a$).

• Convention 1: Upward is positive.
  • An object moving up has $v>0$.
  • An object moving down has $v<0$.
  • Acceleration is always downwards, so $a=-g$.
• Convention 2: Downward is positive.
  • An object moving up has $v<0$.
  • An object moving down has $v>0$.
  • Acceleration is always downwards, so $a=+g$.

A poor choice of sign convention is the most frequent source of errors in these problems.

The following video provides an excellent overview of free-fall motion, including the crucial property of symmetry in upward throws.

Analyzing Free-Fall Scenarios

Let's analyze the two primary free-fall scenarios using our kinematic equations.

Scenario A: Object Thrown Vertically Upwards

Consider an object thrown upwards with an initial velocity $u$. Let's adopt the convention that upward is positive, so $a=-g$.

The video below derives key results for this scenario and highlights the powerful concept of symmetry.

Key properties derived from this analysis:

1. Velocity at Maximum Height: The instantaneous velocity at the peak of the trajectory is $v=0$. The acceleration, however, is still $a=-g$.
2. Maximum Height ($h_{max}$): Using $v^2=u^2+2as$, with $v=0$, $a=-g$, and $s=h_{max}$, we find:
   $0^2=u^2+2(-g)h_{max}⟹h_{max}=\frac{u^2}{2g}$
3. Time of Flight ($T$): This is the total time to return to the initial height. Using $s=ut+\frac{1}{2}a{t}^2$, with $s=0$ and $a=-g$:
   $0=ut-\frac{1}{2}g{t}^2=t\left(u-\frac{1}{2}gt\right)$
   This gives two solutions: $t=0$ (the start) and $T=\frac{2u}{g}$ (the end).
4. Symmetry: The time to reach maximum height (time of ascent) is $t_{ascent}=u/g$, which is exactly half the total time of flight. This implies time of ascent = time of descent. Furthermore, the speed of the object at any given height is the same on the way up as it is on the way down. This symmetry is a powerful problem-solving tool.

Scenario B: Object Dropped from a Height

Consider an object dropped from a height $h$. "Dropped" implies the initial velocity is $u=0$. Here, it's often convenient to set downward as positive. So, $a=+g$.

The video you just watched also includes several worked examples applying these formulas and the concept of symmetry. These are excellent practice for the JEE-style questions.

2. General Strategy for Multi-Stage Motion

Many problems involve motion that changes over time. For example, a car accelerates, then moves at a constant velocity, then brakes. A parachutist is in free-fall, then opens their parachute and decelerates. These are called multi-stage or piecewise motion problems.

The key is that our kinematic equations are only valid for constant acceleration. Therefore, we must break the problem down into stages, where the acceleration is constant within each stage.

The Systematic Approach:

1. Visualize and Diagram: Draw a schematic of the entire motion. Label the start and end points of each stage (e.g., A $\to$ B, B $\to$ C).
2. Establish a Single Coordinate System: Choose an origin and a positive direction. This system must be used for the entire problem.
3. Isolate Each Stage: For each stage, create a list of the five kinematic variables: $\Delta x$, $u$, $v$, $a$, $t$. Fill in the known values.
4. Connect the Stages: The crucial link is that the final velocity and final position of one stage become the initial velocity and initial position of the next stage.
   • $v_{A\to B}=u_{B\to C}$
   • $x_{end,A\to B}=x_{start,B\to C}$
5. Solve for Each Stage: Apply the kinematic equations to each stage to find the unknowns. You will often need to solve for a variable in one stage to use it in the next.

Two-Object Problems: A Special Case of Multi-Stage Motion

A very common problem type involves two objects moving simultaneously. The "stages" here are the individual motions of each object. The strategy is similar:

1. Write down the position equation $x(t)$ for each object: $x_1(t)$ and $x_2(t)$.
2. The linking condition is usually about when and where they meet or collide. At that instant:
   • They are at the same position: $x_1(t)=x_2(t)$.
   • This occurs at the same time $t$.
3. Setting the position equations equal gives you an equation that can be solved for the time $t$ of the event.

The following video masterfully explains this process for a classic two-object free-fall problem.

Worked Example Study

Let's look at a classic multi-stage problem involving a change in acceleration.

Conclusion

In this lesson, you have moved from the theoretical derivation of the kinematic equations to their practical application in complex scenarios. These skills are the bedrock of kinematics problem-solving for the JEE.

Key Takeaways:

• Vertical free-fall is motion with a constant downward acceleration $g$. Your choice of sign convention is the most critical first step in solving any free-fall problem.
• Problems involving an object thrown upwards possess a powerful symmetry: time of ascent equals time of descent, and the speed at any height is the same on the way up and down.
• Multi-stage motion problems are solved by breaking the motion into segments, each with its own constant acceleration. The final state ($v,x$) of one segment serves as the initial state ($u,x_0$) for the next.
• For problems with two moving objects, the key condition is often setting their position functions equal, $x_1(t)=x_2(t)$, to find the time of their meeting.

Preview of the Next Lesson:

We have now thoroughly explored the special case of constant acceleration. But what if acceleration itself changes with time? In our next lesson, we will address the learning outcome: Solve problems with time-dependent acceleration, a(t), using direct integration. This will take us back to the fundamental calculus relationships, $dv=a(t)dt$ and $dx=v(t)dt$, reinforcing the core calculus-based framework of kinematics.