Question

28
If it takes 1,697 m to stop a car initially moving at 46 m/s, what distance is required to stop a car moving at a quarter
of the initial speed under the same conditions?

Answer 1

Answer:

Approximately $106\; \rm m$, $(1/16)$ of the original distance, assuming that the acceleration is constant.

Explanation:

• Let $v$ and $u$ denote the final and initial velocity of the vehicle ($v = 0\; \rm m \cdot s^{-1}$ since this vehicle has come to a stop.)
• Let $a$ denote the acceleration of the vehicle (a constant under the assumption.)
• Let $x$ denote the displacement of the vehicle.

If the acceleration on this vehicle is constant, the SUVAT equations would apply:

$\displaystyle v^{2} - u^{2} = 2\, a\, x$.

Rearrange this equation to find an expression for $x$, the displacement required for the vehicle to come to a stop:

$\begin{aligned} x &= \frac{v^{2} - u^{2}}{2\, a} \\ &= \frac{-u^{2}}{2\, a} && (\text{ v = 0\; \rm m \cdot s^{-1} })\end{aligned}$.

Thus, under these assumptions, $x$ would be proportional to $u^{2}$. In other words, the distance required for this vehicle to come to a stop would be proportional to the square of the initial velocity of this vehicle.

If the initial velocity $u$ is reduced to $(1/4)$ (a quarter) of the initial value, the distance required for stopping this vehicle would be $(1/4)^{2} = (1/16)$ of the initial value:

$\begin{aligned}1,\!697\; {\rm m} \times \frac{1}{16} \approx 106\; \rm m\end{aligned}$.