凌晨3点是什么时辰

Question

Let's say there is a car in front of mine on a straight motorway and its speedometer is showing 100 mph while mine is showing 80mph. But it's decelerating at 10mph/s, while mine is decelerating at 5mph/s. Would the distance between the cars get smaller or larger?

Answer 1

Area under a speed against time graph is the distance travelled.
Car $A$ starts at $100\,\rm mph$ and accelerates at $-10\rm \, mph/s$ and Car $b$ starts at $80\,\rm mph$ and accelerates at $-5\rm \, mph/s$.

.

$\bf 0\le time < 4 \,s$
Car $A$ is moving further away from car $B$.

$\bf 4\,s < time$
Car $B$ is moving closer to car $A$.

---

Algebraically, the speed of car $A$ relative to car $B$ is $(100-10\,t)- (80-5\,t)= 20-5\,t$.
This is positive (ie car $A$ is moving further away from car $B$) when $t<4$ and negative (ie car $B$ is moving closer to car $A$) when $t>4$

Answer 2

Your speed starts at $80$ mph but reduces by $5$ mph per - what ? I am going to guess this is meant to be a deceleration of $5$ mph per second. So after $t$ seconds your speed is $u = 80 - 5t$. And after $t$ seconds the other car's speed is $v = 100 - 10t$.

At time $t=0$ then $v=100$ and $u=80$ so $v>u$ and the distance between the cars is increasing. But at $t=4$ seconds we have $v=60$ and $u=60$, so the speed of the cars is the same. And at a time $t > 4$ seconds we have $u > v$ so the distance between the cars is now decreasing.

Answer 3

Assuming both drivers started to decelerate at the same time. Other notations,- if you drive at speed $u$ before deceleration, then the other driver is going at $u \cdot p$,- some speed greater than yours by factor $p$. Same for deceleration,- other driver starts to decelerate at $a \cdot q$,- some deceleration by factor $q$ greater than yours.

Under these conditions we can construct kinematics equations, your covered distance $s_1$ and other driver's traveled distance $s_2$ : $$ s_1 = ut-1/2at^2 \\ s_2 = upt - 1/2aqt^2 \tag 1$$

Now subtracting one equation from the other in equation system (1) we get how distance between cars evolves over time :

$$ \tag 2 s_2-s_1 = \left( upt-1/2aqt^2 \right)-\left( ut-1/2at^2 \right) $$

Expanding (2) equation and grouping similar terms we arrive at final distance between drivers form :

$$ \tag 3 \Delta s = ut(p-1)-1/2at^2(q-1) $$

Separation of drivers with some $p,q$ factors when drawn in Desmos chart looks like this :
(your car unit speed and deceleration are used in a chart, i.e. $u=1~\text{m/s},~a=1~\text{m/s}^{\text{2}}$)

So, with your problem definition at first distance between drivers increases, but there is always a "no return point" in time when distance between drivers starts to decrease until it reaches zero. This is guaranteed by condition that $aq \gt a.$

EDIT

I have forgot to include initial distance between drivers $s_0$ in the $s_2$ expression of equation system (1). Doing that will not change much separation dynamics $\Delta s (t)$ over time given in Desmos chart. Only that zero distance between drivers will be reached later in time. That's the only profound difference while including $s_0$.