Q.A car moving along a straight highway with speed of $126\ \text{km h}^{-1}$ is brought to a stop within a distance of $200\ \text{m}$. What is the retardation of the car (assumed uniform), and how long does it take for the car to stop?

This problem involves Uniform Acceleration Kinematics, where the car's acceleration (retardation) is constant.

1. First, convert the initial speed to standard SI units:

   $u = 126\ \text{km h}^{-1} = 126 \times \frac{1000\ \text{m}}{3600\ \text{s}} = 126 \times \frac{5}{18}\ \text{m s}^{-1} = 35\ \text{m s}^{-1}$.

   The final speed is $v = 0\ \text{m s}^{-1}$ and the displacement is $s = 200\ \text{m}$.

2. To find the acceleration $a$, use the kinematic equation $v^2 = u^2 + 2as$:

   $0^2 = (35)^2 + 2a(200)$

   $0 = 1225 + 400a$

   $a = -\frac{1225}{400} = -\frac{49}{16}\ \text{m s}^{-2}$.

   The retardation is the magnitude of this acceleration.

3. To find the time $t$ taken to stop, use the kinematic equation $v = u + at$: …

When an object moves along a straight line with a constant acceleration, its motion is described by a set of equations known as the equations of uniform acceleration kinematics. This problem fits perfectly into this framework because the car is moving along a straight highway and its retardation (which is just negative acceleration) is assumed to be uniform (constant).

The core idea is to relate the initial velocity, final velocity, displacement, acceleration, and time using these equations. We'll need to be careful with units and the signs of our quantities. Since the car is slowing down, its acceleration will be in the opposite direction to its initial velocity. If we take the initial direction of motion as positive, then the acceleration will be negative. The term "retardation" specifically refers to the magnitude of this negative acceleration.

Here are the key kinematic equations we might use:

Let's break down the problem step-by-step:

1. Convert Units to SI System

   The initial speed is given in $\text{km h}^{-1}$ and the distance in meters. For consistency in calculations, it's essential to convert all quantities to the standard SI units (meters, seconds).

   The initial speed $u = 126\ \text{km h}^{-1}$.

   To convert $\text{km h}^{-1}$ to $\text{m s}^{-1}$, we multiply by $\frac{1000\ \text{m}}{1\ \text{km}}$ and $\frac{1\ \text{h}}{3600\ \text{s}}$.

   $u = 126\ \text{km h}^{-1} \times \frac{1000\ \text{m}}{1\ \text{km}} \times \frac{1\ \text{h}}{3600\ \text{s}}$

   $u = 126 \times \frac{1000}{3600}\ \text{m s}^{-1}$

   $u = 126 \times \frac{5}{18}\ \text{m s}^{-1}$

   $u = 7 \times 5\ \text{m s}^{-1}$

   $u = 35\ \text{m s}^{-1}$

   Watch out

   Unit conversion is a common source of errors. Always ensure all quantities are in a consistent system (like SI units) before performing calculations.

2. Identify Knowns and Unknowns

   Let's list the given information and what we need to find:

   • Initial velocity, $u = 35\ \text{m s}^{-1}$ (positive, assuming the direction of motion is positive).
   • Final velocity, $v = 0\ \text{m s}^{-1}$ (the car comes to a stop).
   • Displacement, $s = 200\ \text{m}$ (positive, in the direction of motion).
   • Acceleration, $a = ?$ (This will be negative, representing retardation).
   • Time, $t = ?$

3. Calculate the Retardation ($a$)

   We need an equation that relates $u, v, s,$ and $a$. The equation $v^2 = u^2 + 2as$ is perfect for this, as it does not involve time $t$, which is currently unknown.

   Substitute the known values:

   $0^2 = (35\ \text{m s}^{-1})^2 + 2 \times a \times (200\ \text{m})$

   $0 = 1225\ \text{m}^2\text{s}^{-2} + 400a\ \text{m}$

   Rearrange to solve for $a$:

   $400a = -1225$ …

Concept: Finding Time Before Acceleration, from the v-t Graph's Area

Method: Area-First Approach (get $t$ directly from displacement, then $a$ from the slope)

The usual route finds the deceleration $a$ first (via $v^2=u^2+2as$) and then the time $t$ (via $v=u+at$). This method reverses that order: since displacement is the area under the $v$-$t$ graph, and that area depends on $u$, $s$ and $t$ but not directly on $a$, the stopping time can be extracted first — before acceleration is ever computed.

Steps

1. Convert the initial speed to SI units:

2. Sketch the $v$-$t$ graph. It is a straight line from $v=35\ \text{m s}^{-1}$ (at $t=0$) down to $v=0$ (at the unknown stopping time $t$) — a right triangle with base $t$ and height $u=35\ \text{m s}^{-1}$.

3. Displacement = area of that triangle. This relation only involves $u$, $s$, and $t$ — no acceleration term appears:

4. Solve directly for $t$, using the given $s = 200\ \text{m}$:

Notice: the retardation was never needed to reach this result.

5. Now find the retardation from the slope of the same line — magnitude of slope $=$ (change in speed) / (time):

Why finding $t$ first is a genuine shortcut here …