Animated Solution for Mathematics - Differentiation: Two points A and B move from rest along a straight line with constant acceleration f and f' respectively. If A takes m sec. more than B and describes n units more than B in acquiring the same speed then

Welcome, future engineers! Today, we are not just solving a kinematics problem; we are dissecting the very geometry of motion. Imagine two particles, $A$ and $B$, standing at the starting line.

They are about to embark on a journey where they accelerate from rest to a common final velocity, $v$. They do so with different accelerations, $f$ and $f^{\prime }$. This is a classic JEE Advanced scenario where the algebra can get messy if you dive in blindly, but if you look at it through the lens of a velocity-time graph, the complexity melts away.

The Velocity-Time Landscape

Let us visualize this. On a velocity-time graph, the vertical axis is velocity, and the horizontal axis is time. Both particles start at the origin $(0,0)$.

Since they accelerate constantly, their paths are straight lines. The slope of these lines represents the acceleration. We are told that particle $A$ takes $m$ seconds more than $B$ to reach the same speed $v$.

This means $A$'s line is 'gentler'—it has a smaller slope, $f$, while $B$'s line is steeper, with slope $f^{\prime }$. This geometric intuition is our first victory.

The Algebra of Time

Now, let us translate this into the language of equations. For any object starting from rest with constant acceleration $a$, the first equation of motion is $v=u+at$. Since $u=0$, we have $v=at$, or $t=\frac{v}{a}$.

For particle $A$, the time taken is $t_A=\frac{v}{f}$. For particle $B$, the time taken is $t_B=\frac{v}{f^{\prime }}$.

We are given that $A$ takes $m$ seconds more than $B$, so $t_A-t_B=m$. Substituting our expressions, we get:

Factoring out $v$, we arrive at our first pillar of truth:

The Geometry of Distance

Next, we look at the distance covered. In a velocity-time graph, the distance is the area under the line. For a triangle with base $t$ and height $v$, the area is:

Since $t=\frac{v}{a}$, the distance is $s=\frac{v^2}{2a}$. Thus, the distance for $A$ is $s_A=\frac{v^2}{2f}$, and for $B$, it is $s_B=\frac{v^2}{2f^{\prime }}$.

We are told $A$ covers $n$ units more than $B$, so $s_A-s_B=n$. Substituting our expressions:

Factoring out $\frac{v^2}{2}$, we get our second pillar:

The Elegant Cancellation

Here is where the magic happens. Look at Equation 1 and Equation 2. They share the exact same term: $\frac{f^{\prime }-f}{f{f}^{\prime }}$.

This is not a coincidence; it is the symmetry of the problem. If we divide Equation 2 by Equation 1, that entire complex term vanishes:

This simplifies to $\frac{v}{2}=\frac{n}{m}$, which gives us the final velocity:

We are almost home. Substitute $v=\frac{2n}{m}$ back into Equation 1:

Multiply both sides by $m$ and rearrange:

And there it is! We have arrived at the solution. This result is not just a collection of variables; it is a testament to the power of looking for symmetry in physics. When you face these problems in the exam, do not just calculate—observe. The math will always reward you for it.