Two cars at the traffic light

Problem

Car A passes a green traffic light with constant velocity $v_A = 20\;\mathrm{m/s}$. At the same instant a car B, initially at rest at the same traffic light, sets off with constant acceleration $a_B = 4{,}0\;\mathrm{m/s}^2$. Determine after how much time B catches up with A and at which position. Is this a realistic situation in an urban context?

Solution

Setup. We place the origin at the traffic light and start the clock at the instant A passes and B sets off. We write the equations of motion of the two cars, measuring position along the road:

$x_A(t) = v_A\,t = 20\,t, \qquad x_B(t) = \tfrac{1}{2}\,a_B\,t^2 = 2\,t^2$

Meeting condition. B catches up with A when the two positions coincide, $x_A = x_B$:

$20\,t = 2\,t^2 \quad \Rightarrow \quad 2\,t^2 - 20\,t = 0 \quad \Rightarrow \quad 2t\,(t - 10) = 0$

The solutions are $t = 0$ (the starting instant, when they are side by side at the traffic light) and the physically interesting solution:

$t = 10\;\mathrm{s}$

Meeting position. Substituting into either equation of motion:

$d = x_A(10) = 20 \cdot 10 = 200\;\mathrm{m}$

Discussion of realism. At the moment of overtaking, B’s velocity is $v_B = a_B\,t = 4{,}0 \cdot 10 = 40\;\mathrm{m/s}$, that is $40 \cdot 3{,}6 = 144\;\mathrm{km/h}$. An acceleration of $4{,}0\;\mathrm{m/s}^2$ sustained for $10\;\mathrm{s}$ is unrealistic in a city: no ordinary car sustains such an acceleration for that long, and in any case $144\;\mathrm{km/h}$ in a built-up area is well beyond the speed limits. The constant-acceleration model therefore only holds for the first few instants.

$\ev{\Delta t = 10\;\mathrm{s}, \quad d = 200\;\mathrm{m}}$

Collegamenti

Argomenti: Cinematica Concetti: Moto uniformemente accelerato