At the initial instant, two particles are observed at different locations moving towards each other

Q: Why do they meet twice?

Because their accelerations are opposite to their initial velocities. They start moving towards each other, meet, cross each other, but then they are constantly slowing down. Eventually, they stop and reverse direction, causing them to meet a second time.

Q: Can I solve this without relative motion?

Yes, you can write $x_1 = u_1 t - \frac{1}{2} a_1 t^2$ and $x_2 = d - u_2 t + \frac{1}{2} a_2 t^2$, and equate them. You will end up with the exact same quadratic equation, but relative motion makes the setup conceptually cleaner.

Q: Why is relative acceleration $a_1 + a_2$?

Acceleration of particle 1 is $-a_1 \hat{i}$ and particle 2 is $a_2 \hat{i}$. Relative acceleration of 1 w.r.t 2 is $a_{12} = a_1 - a_2 = -a_1 \hat{i} - a_2 \hat{i} = -(a_1 + a_2) \hat{i}$. The magnitude is $a_1 + a_2$, directed opposite to their relative approach velocity.

Q: What if the roots of the quadratic are imaginary?

If the roots are imaginary, it means the particles never meet. They slow down and reverse direction before their paths cross. The problem states they *do* meet twice, so the discriminant must be strictly positive.

Q: How did we get the formula for $(\Delta t)^2$?

It's a standard algebraic identity: $(a - b)^2 = (a + b)^2 - 4ab$. We used this to relate the difference of the roots (which is $\Delta t$) to the sum and product of the roots, which we can easily read from the quadratic equation.

Question

At the initial instant, two particles are observed at different locations moving towards each other with velocities $u_1$ and $u_2$. If they are subjected to constant accelerations $a_1$ and $a_2$ in directions opposite to their initial velocities, they will meet twice. If time interval between these two meetings is $\Delta t$, find suitable expression for their initial separation.

Sigma Percentile · Accepted Answer

The beauty of physics often lies in its ability to take a complex, multi-body problem and reduce it to a single, elegant equation. This problem is a classic example of how choosing the right frame of reference can turn a potentially messy algebraic nightmare into a straightforward quadratic equation.

### The Physical Setup: A Tale of Two Particles

Imagine you are standing on a straight road, watching two cars, A and B, initially separated by a distance $d$. They are driving towards each other with initial speeds $u_1$ and $u_2$.

However, there is a twist. Both drivers suddenly hit their brakes! This means they are subjected to constant accelerations $a_1$ and $a_2$ that are directed **opposite** to their initial velocities.

Because they are decelerating, they might not crash. In fact, the problem tells us they meet **twice**. How is this physically possible?

They meet the first time while they are still moving towards each other. After crossing paths, they continue to slow down until they momentarily come to a halt. Then, because the acceleration is constant and still acting, they reverse their directions. As they move back towards their starting points, they cross paths a second time!

### The Power of Relative Motion

We could solve this by writing the position-time equations for both particles from the ground frame and equating them. But there is a much more powerful tool at our disposal: **Relative Motion**.

Let's hop into the driver's seat of car B. From B's perspective, B is completely stationary. What does car A look like?

Car A is coming towards B with a relative velocity $u_{	ext{rel}} = u_1 + u_2$.

What about the acceleration? Car A is decelerating at $a_1$ (away from B), and car B is decelerating at $a_2$ (away from A). In B's frame, A appears to have a relative acceleration $a_{	ext{rel}} = a_1 + a_2$, directed opposite to its relative velocity.

### The Master Equation: Quadratic in Time

Now, the problem is reduced to a single particle (A) moving towards a stationary point (B) from an initial distance $d$.

When they meet, the relative displacement of A must be exactly $d$. We can use the second equation of motion:

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

Substituting our relative variables, we get:

$$d = (u_1 + u_2)t - \frac{1}{2}(a_1 + a_2)t^2$$

Notice the negative sign! It is crucial because the relative acceleration opposes the relative velocity.

Let's rearrange this into a standard quadratic equation in terms of time $t$:

$$\frac{1}{2}(a_1 + a_2)t^2 - (u_1 + u_2)t + d = 0$$

This beautiful equation holds the key to the entire problem. Because it is a quadratic, it will generally have two roots, $t_1$ and $t_2$. These roots represent the exact two moments in time when the particles meet!

### Unlocking the Roots

We are given that the time interval between these two meetings is $\Delta t$. Mathematically, this means $\Delta t = |t_2 - t_1|$.

To connect this to our quadratic equation, we use the fundamental properties of roots. For a quadratic equation $Ax^2 + Bx + C = 0$, the sum and product of the roots are given by $-B/A$ and $C/A$ respectively.

For our equation, the sum of the roots is:

$$t_1 + t_2 = \frac{2(u_1 + u_2)}{a_1 + a_2}$$

And the product of the roots is:

$$t_1 t_2 = \frac{2d}{a_1 + a_2}$$

### The Final Algebraic Flourish

We need to link the difference of the roots ($\Delta t$) to their sum and product. We use the classic algebraic identity:

$$(t_2 - t_1)^2 = (t_1 + t_2)^2 - 4t_1 t_2$$

Substituting $\Delta t$ and our expressions for the sum and product, we get:

$$(\Delta t)^2 = \left( \frac{2(u_1 + u_2)}{a_1 + a_2} ight)^2 - 4 \left( \frac{2d}{a_1 + a_2} ight)$$

This equation now contains only known variables and our single unknown, the initial separation $d$.

Let's isolate the term containing $d$:

$$\frac{8d}{a_1 + a_2} = \frac{4(u_1 + u_2)^2}{(a_1 + a_2)^2} - (\Delta t)^2$$

Finally, multiplying both sides by $\frac{a_1 + a_2}{8}$, we arrive at our magnificent final answer:

$$d = \frac{(u_1 + u_2)^2}{2(a_1 + a_2)} - \frac{(a_1 + a_2)(\Delta t)^2}{8}$$

This expression perfectly captures the initial separation required for the two particles to meet exactly twice with a time gap of $\Delta t$. It is a testament to how relative motion and basic algebra can elegantly unravel complex kinematic dances.

Visualized Solution

$s_{rel} = d$

$u_{rel}, a_{rel}$

$s = u t + \frac{1}{2} a t^{2}$

$\frac{1}{2} (a_{1} + a_{2}) t^{2} - (u_{1} + u_{2}) t + d = 0$

$t_{1} + t_{2}, t_{1} t_{2}$

$(Δ t)^{2} = (t_{1} + t_{2})^{2} - 4 t_{1} t_{2}$

$d = \frac{( u _{1} + u _{2} ) ^{2}}{2 ( a _{1} + a _{2} )} - \frac{( a _{1} + a _{2} ) ( Δ t ) ^{2}}{8}$

$t_{1} \neq = t_{2}$

The Sigma Insight: Relative Velocity

The Physical Setup

The Power of Relative Motion

The Master Equation

Unlocking the Roots

The Final Algebraic Flourish

Similar Questions

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Two material particles A and B are moving in free space. How their position coordinates x, y and z vary with time t is shown in the following graphs. Determine at what instant of time the particles are closest to each other and the closest separation.

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } srel​=d

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } urel​,arel​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } s=ut+21​at2

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 21​(a1​+a2​)t2−(u1​+u2​)t+d=0

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } t1​+t2​,t1​t2​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (Δt)2=(t1​+t2​)2−4t1​t2​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } d=2(a1​+a2​)(u1​+u2​)2​−8(a1​+a2​)(Δt)2​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } t1​=t2​