Animated Solution for Physics - Kinematics: A car starting from rest, accelerates at the rate $f$ through a distance $s$, then continues at constant speed for time $t$ and then decelerates at the rate $f/2$ to come to rest. If the total distance travelled is $15s$, then

The beauty of physics often lies in choosing the right tool for the job. When dealing with multiple phases of motion—acceleration, constant velocity, and deceleration—relying purely on algebraic kinematic equations like $v=u+at$ and $s=ut+\frac{1}{2}a{t}^2$ can quickly turn into a tangled web of variables.

Instead, we can unlock the solution elegantly by visualizing the journey through a Velocity-Time ($v-t$) graph. Let's embark on this geometric journey to decode the car's motion.

The Three Phases of Motion

Imagine the car's journey plotted on a graph where the vertical axis is velocity ($v$) and the horizontal axis is time ($t$). The motion is split into three distinct phases:

Phase 1: The Acceleration The car starts from rest ($v=0$ at $t=0$) and accelerates at a constant rate $f$. On our graph, this is a straight line shooting upwards from the origin. The slope of this line is $f$. Let's say this phase lasts for a time $t_1$. The maximum velocity reached is $v_{max}=f{t}_1$. The distance covered, $s$, is simply the area of the triangle formed under this line:

Phase 2: The Cruise The car now maintains this maximum velocity $v_{max}$ for a given time $t$. This forms a flat, horizontal rectangle on our graph. The distance covered during this cruise is the area of the rectangle:

Phase 3: The Deceleration Finally, the brakes are hit. The car decelerates at a rate of $f/2$ until it comes to a complete halt. This forms a downward sloping line. Let's call the time taken to stop $t_2$. The slope here is $-f/2$.

Geometry of the V-T Graph

Here is where the magic happens. We can express the peak velocity $v_{max}$ using both the acceleration and deceleration phases:

Equating the two gives us a beautiful relationship:

This makes perfect intuitive sense: if you brake half as hard as you accelerated, it will take you twice as long to stop!

Now, what about the distance covered during braking? Let's calculate the area of that final triangle ($s_3$):

Notice something fascinating? Since $s=\frac{1}{2}f{t}_1^2$, the braking distance $s_3$ is exactly double the initial acceleration distance. So, $s_3=2s$.

Bringing it all together

The problem hands us a golden key: the total distance traveled across all three phases is $15s$. Let's sum up our geometric areas:

We are almost at the finish line. Let's substitute our expressions for $v_{max}$ and $s$ back into this equation to find a relationship between the times:

Assuming $f eq 0$ and $t_1 eq 0$, we can divide both sides by $f{t}_1$:

To find the final answer, we substitute $t_1$ back into our original equation for $s$:

Looking at the given options, $\frac{1}{72}f{t}^2$ is nowhere to be found. Therefore, the correct choice is confidently (d) None of the above.