Animated Solution for Physics - Kinematics: A car accelerates from rest at a constant rate $\alpha$ for some time after which it decelerates at a constant rate $\beta$ to come to rest. If the total time elapsed is $t$ seconds, the total distance travelled is

Imagine you are driving a car on a long, straight highway. You press the gas pedal, and the car accelerates at a constant rate $\alpha$. You reach a top speed, and then immediately hit the brakes, decelerating at a constant rate $\beta$ until you come to a complete stop. The entire journey takes exactly $t$ seconds. Our goal is to find the total distance you travelled during this time.

While we could use the standard equations of motion ($v=u+at$ and $s=ut+\frac{1}{2}a{t}^2$) to solve this, there is a much more elegant and powerful method: Graphical Analysis.

Let's plot the car's velocity against time.

Since the car starts from rest, the graph begins at the origin $(0,0)$. It accelerates uniformly, meaning the velocity increases linearly, forming a straight line with a positive slope $\alpha$. Let's say it reaches a maximum velocity $v_0$ at time $t_1$.

Then, the car decelerates uniformly to rest. The graph goes down as a straight line with a negative slope $-\beta$, hitting the time axis at the total time $t$. Let the duration of this deceleration phase be $t_2$.

The slope of a velocity-time graph represents acceleration.

For the acceleration phase, the slope is $\alpha$:

For the deceleration phase, the magnitude of the slope is $\beta$:

We know the total time of the journey is $t$. Therefore, the sum of the acceleration time and deceleration time must equal $t$:

Substituting our expressions for $t_1$ and $t_2$:

Now, let's factor out the maximum velocity $v_0$:

Finding a common denominator gives us:

Rearranging this to solve for $v_0$, we get a beautiful standard result for the peak velocity:

To find the total distance travelled, we need to find the area under the velocity-time graph. Because the car never reverses direction, the displacement and distance are identical.

Our graph forms a simple triangle with a base of $t$ and a height of $v_0$. The area of a triangle is:

Finally, we substitute the expression for $v_0$ that we just derived:

Multiplying the terms together, we arrive at our final answer:

This is a classic and highly useful formula in kinematics. Memorizing it can save you valuable time in competitive exams!