Problem — Challenge: catching the bus

Problem

Challenge: catching the bus. A bus starts from rest with $a = 1{,}2\;\text{m/s}^2$. You are $8\;\text{m}$ behind the door and start running. What is the minimum average speed needed to catch it before the interception window closes? (Hint: impose a non-negative discriminant in the interception equation.)

Solution

Setup. We place the origin at the bus door at the starting instant. The door moves with uniformly accelerated motion, while you run at average speed $\bar v$ starting $8\;\text{m}$ behind:

$x_\text{bus}(t) = \frac{1}{2}\,a\,t^2 = 0{,}6\,t^2, \qquad x_\text{tu}(t) = \bar v\,t - 8$

Interception condition. You reach the door when $x_\text{tu} = x_\text{bus}$:

$\bar v\,t - 8 = 0{,}6\,t^2 \;\Rightarrow\; 0{,}6\,t^2 - \bar v\,t + 8 = 0$

Existence of the solution. For a real meeting instant to exist, the discriminant of this second-degree equation in $t$ must be non-negative:

$\Delta = \bar v^2 - 4\cdot 0{,}6\cdot 8 \geq 0 \;\Rightarrow\; \bar v^2 \geq 19{,}2$

Minimum speed. The limiting case $\Delta = 0$ gives the smallest average speed with which you can still catch the bus (at a single instant of tangency):

$\bar v_\text{min} = \sqrt{19{,}2} \approx 4{,}38\;\text{m/s}$

$\ev{\bar v_\text{min} \approx 4{,}38\;\mathrm{m/s} \approx 15{,}8\;\mathrm{km/h}}$

At a lower speed ($\Delta < 0$) there is no instant at which you reach the door: the bus, accelerating, escapes you forever. Almost $16\;\text{km/h}$ is a sustained running pace: catching the bus on the fly is harder than it looks.

Links

Topics: Cinematica Concepts: Moto uniformemente accelerato