Animated Solution for Physics - Kinematics: In a car race on a straight path, car A takes a time t less than car B at the finish and passes finishing point with a speed 'v' more than that of car B. Both the cars start from rest and travel with constant acceleration $a_1$ and $a_2$ respectively. Then 'v' is equal to

Visualizing the Race

Imagine the roar of engines at the starting line of a drag race. We have two cars, Car A and Car B, both waiting for the green light. They both start from rest, meaning their initial velocities are zero ($u=0$). However, they are not identical machines. Car A has a constant acceleration of $a_1$, and Car B has a constant acceleration of $a_2$.

The problem tells us that Car A is the faster of the two. It crosses the finish line a time $t$ before Car B does. To translate this into math, let's assign a variable. Let the time taken by Car A to finish the race be $t_0$. Because Car B is slower and takes time $t$ longer, the time taken by Car B to cross that exact same finish line is $t_0+t$.

Furthermore, when they cross the finish line, they are moving at different speeds. Car A crosses the line with a velocity $v_A$, and Car B crosses with a velocity $v_B$. The problem states that Car A's finishing speed is $v$ greater than Car B's. Mathematically, this means $v_A-v_B=v$. Our ultimate goal is to find an expression for this velocity difference $v$ in terms of the given variables $a_1$, $a_2$, and $t$.

To solve this, we need to reach into our physics toolbox and pull out the standard equations of kinematics for constant acceleration. Since we are dealing with initial velocity, final velocity, acceleration, time, and distance, two equations perfectly fit our needs:

1. The velocity-time relation: $v=u+at$ 2. The displacement-time relation: $s=ut+\frac{1}{2}a{t}^2$

Since both cars start from rest, $u=0$, these equations simplify beautifully to:

Let's apply these simplified equations to both cars at the moment they cross the finish line.

For Car A: Final velocity: $v_A=a_1{t}_0$ Distance covered: $s_A=\frac{1}{2}{a}_1{t}_0^2$

For Car B: Final velocity: $v_B=a_2(t_0+t)$ Distance covered: $s_B=\frac{1}{2}{a}_2(t_0+t{)}^2$

We know the velocity difference is $v$. Substituting our expressions for $v_A$ and $v_B$, we get our first master equation:

This equation is almost what we need, but it contains $t_0$, which is an unknown variable we introduced. We need to eliminate it.

Equating Distances

How do we find $t_0$? We use the one piece of physical reality we haven't exploited yet: both cars raced on the exact same track. Therefore, the distance covered by Car A is exactly equal to the distance covered by Car B.

Substituting our distance expressions:

The $\frac{1}{2}$ on both sides cancels out immediately.

Now, we face a choice. We could expand the right side and solve a messy quadratic equation for $t_0$. But there is a much more elegant path. Since time and acceleration are positive quantities, we can simply take the square root of both sides! This is a crucial algebraic maneuver that saves us a lot of time.

Now, let's isolate $t_0$. First, expand the right side:

Bring all terms containing $t_0$ to the left side:

Finally, divide to solve for $t_0$:

Calculating Velocity Difference

We have successfully expressed our unknown $t_0$ entirely in terms of the known variables $a_1$, $a_2$, and $t$. The final phase of our solution is to substitute Equation 2 back into Equation 1.

Recall Equation 1:

Substitute $t_0$:

At first glance, this looks complicated. But look closely at the term $(a_1-a_2)$. We can use the difference of squares algebraic identity, $x^2-y^2=(x-y)(x+y)$, in reverse. We can treat $a_1$ as $(\sqrt{a_1}{)}^2$ and $a_2$ as $(\sqrt{a_2}{)}^2$.

Let's substitute this expanded form into our velocity equation:

The beauty of physics and math combined! The problematic denominator $(\sqrt{a_1}-\sqrt{a_2})$ perfectly cancels out with the term in the numerator.

Now, simply distribute the $\sqrt{a_2}t$ into the parenthesis:

The $+a_{2}t$ and $-a_{2}t$ terms annihilate each other, leaving us with a remarkably clean and elegant final result:

Through a systematic application of basic kinematic equations and a clever algebraic substitution, we've navigated from a complex physical scenario to a beautifully simple mathematical truth. This problem is a classic example of why mastering algebraic manipulation—like recognizing a hidden difference of squares—is just as important as understanding the underlying physics principles. The final expression, $v=\sqrt{a_1{a}_2}t$, shows that the velocity difference is directly proportional to the time difference and the geometric mean of their accelerations.