A body travels a distance s in t seconds. It starts from rest and ends at rest. In the first part of

Q: Wait, why can't we just use the kinematic equation $v^2 = u^2 + 2as$ for the whole journey?

That is a classic trap! The equation $v^2 = u^2 + 2as$ assumes a constant acceleration throughout the entire displacement. Here, the acceleration changes abruptly from $f$ to $-r$ halfway through. You must treat the two phases separately or use the graphical method.

Q: Is it always a triangle? What if the acceleration wasn't constant?

Great observation! If the acceleration were not constant, the velocity-time graph would be a curve, and the area would require integration. Because the problem specifies 'constant' acceleration and retardation, the graph is guaranteed to be linear, forming a perfect triangle.

Q: Why did we define $v$ as the maximum velocity?

Because the body accelerates from rest to a peak velocity $v$ and then decelerates back to rest. The peak of the triangle in our $v-t$ graph represents the point where the motion switches from acceleration to retardation.

Q: Can I solve this using $s = ut + \frac{1}{2}at^2$ for both parts?

Absolutely! You can write $s_1 = \frac{1}{2}ft_1^2$ and $s_2 = v t_2 - \frac{1}{2}rt_2^2$. It is mathematically equivalent, but it involves more algebraic steps. The graphical method is much faster and less prone to calculation errors.

Q: Does the order of $f$ and $r$ matter in the final formula?

Look closely at the term $(\frac{1}{f} + \frac{1}{r})$. Addition is commutative, so $f+r$ or $r+f$ doesn't change the result. The physics remains symmetric regardless of which phase happens first!

Question

A body travels a distance s in t seconds. It starts from rest and ends at rest. In the first part of the journey, it moves with constant acceleration f and in the second part with constant retardation r. The value of t is given by

Options:

(A) $\sqrt{2s(\frac{1}{f} + \frac{1}{r})}$

(B) $2s(\frac{1}{f} + \frac{1}{r})$

(C) $\frac{1}{f} + \frac{1}{r}$

(D) $\sqrt{2s(f + r)}$

Sigma Percentile · Accepted Answer

### The Geometry of Motion: A Journey from Rest to Rest

Imagine you are standing at a starting line. You begin to run, accelerating at a constant rate $f$ until you hit your peak speed.

Then, realizing you need to stop exactly at a finish line a distance $s$ away, you begin to decelerate at a constant rate $r$. This is the essence of our problem. It is not just a collection of variables; it is a story of momentum, control, and the elegant symmetry of kinematics.

### Phase 1: The Velocity-Time Canvas

When we face a problem involving constant acceleration and retardation, the most powerful tool in our arsenal is the velocity-time ($v-t$) graph.

Since the body starts at rest ($u=0$) and ends at rest ($v_{final}=0$), our graph starts at the origin $(0,0)$ and returns to the time axis at $t$. The velocity increases linearly to a maximum value $v$ and then decreases linearly back to zero, creating a perfect triangle.

### Phase 2: Decoding the Slopes

In a $v-t$ graph, the slope represents acceleration. For the first phase, the slope is $f$, so we have:

$$f = \frac{v}{t_1} \Rightarrow t_1 = \frac{v}{f}$$

For the second phase, the slope is $-r$ (where $r$ is the magnitude of retardation), so:

$$r = \frac{v}{t_2} \Rightarrow t_2 = \frac{v}{r}$$

The total time for the journey is the sum of these two intervals:

$$t = t_1 + t_2 = \frac{v}{f} + \frac{v}{r} = v \left( \frac{1}{f} + \frac{1}{r} ight)$$

### Phase 3: The Area Under the Curve

We invoke the fundamental principle that the area under a $v-t$ graph equals the total displacement. Our triangle has a base of $t$ and a height of $v$.

The area is given by:

$$s = \frac{1}{2} 	imes t 	imes v$$

Our goal is to eliminate $v$ to express $t$ in terms of $s$, $f$, and $r$.

### The Final Synthesis

From our time equation, we isolate $v$:

$$v = \frac{t}{\left( \frac{1}{f} + \frac{1}{r} ight)}$$

Substituting this into our area equation:

$$s = \frac{1}{2} 	imes t 	imes \left( \frac{t}{\frac{1}{f} + \frac{1}{r}} ight) = \frac{t^2}{2 \left( \frac{1}{f} + \frac{1}{r} ight)}$$

With a quick cross-multiplication, we arrive at the final relationship:

$$t^2 = 2s \left( \frac{1}{f} + \frac{1}{r} ight)$$

Taking the square root, we reach our destination:

**$$t = \sqrt{2s \left( \frac{1}{f} + \frac{1}{r} ight)}$$**

This result is a testament to how geometry can simplify the most daunting physical problems. You have successfully navigated the acceleration, the peak, and the retardation. Keep this clarity of thought, and no JEE problem will ever be too complex to conquer.

Animated Solution for Mathematics - Straight Lines: A body travels a distance s in t seconds. It starts from rest and ends at rest. In the first part of the journey, it moves with constant acceleration f and in the second part with constant retardation r. The value of t is given by

Select Answer:

Visualized Solution

Velocity-Time Graph Setup

Phase 1: Constant Acceleration

Phase 2: Constant Retardation

Relating Slopes to Time

Total Time of Journey

Isolating Maximum Velocity $v$

Distance from $v - t$ Graph

Setting Up the Distance Equation

Eliminating Velocity $v$

Simplifying the Expression

Rearranging for $t^{2}$

Final Expression for Time $t$

The Sigma Insight: Slope of a Line

The Geometry of Motion

Phase 1

Phase 2

Phase 3

The Final Synthesis

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