Two cars are stopped at a light. Car A is in the front. Car B is right behind it. Distance between is zero. Both cars are stopped. The light turns green. Car A accelerates at constant rate of 1. Car B must accelerate such that it is always exactly 3 times its velocity behind car A. What is the function for the velocity of car B at time t? (Car A obviously has velocity t at time t. It also obviously has position 1/2 t squared at time t.)
Or:
stopped . in front. behind , @ distance 0. accelerates at constant rate of 1. accelerates, being exactly 3 times its velocity behind . Solve for velocity of at time t? ( obviously has velocity t at time t.)
Or by specifying position and velocity, can we solve over a non-trivial domain?
Equation of the form: $$v+kv’=t$$
Seems to apply. What is solution?
$$v=t-k+ke^{-t/k}$$
The lead car always moves faster than the following car accept when they are both at rest.
The final question then is if there are $n$ cars, instead of just two, in line, what is the equation of motion for the last in line, car $n$?
Let $s_a$ be the position of car $A$, with $0$ at the start. Let $v_a$ be the speed of car $A$. Let $s_b$ be the position of car $B$, measured from where $B$ starts, which means $s_a=s_b$ means the cars are touching. Let $v_b$ be the speed of car $B$. Let $a_b$ be the acceleration of car $B$.
The speed of $A$ is $v_a=1t$, where the $1$ is the acceleration.
Its position is $s_a=\frac 12t^2$
We want $s_b=s_a-3v_b=\frac 12t^2-3v_b$
This is $3\frac {ds_b}{dt}+s_b=\frac 12t^2$ with $s_b(0)=0$
$s_b=\frac 12t^2-3t+9-9e^{-t/3}$ from Alpha
For more cars, you follow the process down the line. Solving the equation for the second car gives the input for the third car and so on.