Q.Stopping distance of vehicles: When brakes are applied to a moving vehicle, the distance it travels before stopping is called stopping distance. It is an important factor for road safety and depends on the initial velocity ($v_0$) and the braking capacity, or deceleration, $-a$ that is caused by the braking. Derive an expression for stopping distance of a vehicle in terms of $v_0$ and $a$.

Concept: Uniformly Accelerated Motion

A vehicle braking to rest undergoes constant deceleration. We need to relate the stopping distance to initial velocity and deceleration without involving time.

Start with the kinematic equation that connects velocity, acceleration, and displacement:

where $v$ is the final velocity, $v_0$ is the initial velocity, $a$ is the acceleration (negative for braking), and $s$ is the displacement.

At the moment the vehicle stops, $v = 0$. Substituting:

$v_0^2 = 2as$ …

When a driver applies the brakes, the vehicle doesn't stop instantly. The friction between the brake pads and wheels creates a constant retarding force, which produces a uniform deceleration. This is a classic example of uniformly accelerated motion—except here the acceleration is negative, slowing the vehicle down.

The key insight is that we know three things: the initial velocity $v_0$, the final velocity (zero, since the vehicle stops), and the deceleration $a$. We want to find the displacement during this process. The kinematic equation that connects these quantities without involving time is perfect for this situation.

Derivation

1. Identify the known quantities and what we seek.

   The vehicle starts with velocity $v_0$, experiences a constant deceleration of magnitude $a$ (so the acceleration is $-a$ in the direction of motion), and comes to rest, meaning final velocity $v = 0$. We need to find the displacement $s$ during braking—this is the stopping distance.

2. Choose the appropriate kinematic equation.

   For uniformly accelerated motion, we have several equations. Since we don't know (or care about) the time taken to stop, we use the equation that relates velocity, acceleration, and displacement:

Here $a$ represents the acceleration. Since the vehicle is decelerating, we substitute $a = -a$ (negative because it opposes motion).

3. Substitute the final velocity and deceleration.

   At the moment the vehicle stops, $v = 0$. The equation becomes:

4. Solve for stopping distance $s$.

   Rearranging:

Concept: Stopping Distance as the Area Under a v-t Graph

Method: Graphical (Triangle-Area) Derivation

Displacement during any motion equals the area under its velocity-time graph. For a vehicle braking uniformly from $v_0$ to rest, that graph is a single right-angled triangle — its area gives the stopping distance directly, without substituting into the algebraic kinematic equation.

Steps

1. Sketch the $v$-$t$ graph. Speed starts at $v_0$ (at $t=0$) and falls in a straight line (constant deceleration $a$) to $v=0$ at some stopping time $t_s$.

2. Find the stopping time from the slope. The magnitude of the line's slope is the deceleration $a$:

3. Identify the shape under the graph. Between $t=0$ and $t=t_s$, the region under the line is a right triangle with base $t_s$ and height $v_0$.

4. Stopping distance = area of that triangle:

5. Substitute $t_s = v_0/a$ from Step 2: …