Hello, Mr_Green!
Everyone seems to have overlooked the essential piece of data:
. . the lengths of the cars.
A 35-mph car is driving down the street.
It catches up to a slow poke car going 15 mph.
The 35-mph car wants to pass him.
How much room (distance) will he need to pass the 15-mph car? If the cars are considered to be POINTS,
. . it takes only an instant to complete the passing.
Assuming the cars have length $\displaystyle L$ feet,
. . the faster car $\displaystyle (F)$ overtakes the slower car $\displaystyle (S)$
. . when F's front bumper is even with S's rear bumper.
And that is when we begin measuring time and distance. Code:
L
*---------*
| S | →
*---------*
L
*---------*
| F | →
*---------*
At some point down the road, F has passed S.
F's rear bumper is even with S's front bumper.
And F has travelled a distance $\displaystyle d.$ Code:
L
*---------*
| S | →
*---------*
L
*---------*
| F | →
+ - - - - - - d - - - - - - *---------*
Relative to $\displaystyle S$, $\displaystyle F$ must travel a distance of $\displaystyle 2L$ feet.
Relative to $\displaystyle S$, $\displaystyle F$'s speed is: .$\displaystyle 35 - 15 \:=\:20\text{ mph} \:=\:\frac{88}{3}\text{ ft/sec}$
This will take: .$\displaystyle \frac{2L}{\frac{88}{3}} \:=\:\frac{3L}{44}$ seconds.
At $\displaystyle 35\text{ mph} \:=\:\frac{154}{3}\text{ ft/sec, }F$ travels: .$\displaystyle \left(\frac{154}{3}\text{ ft/sec}\right) \times \left(\frac{3L}{44}\text{ sec}\right) \;=\;\frac{7}{2}L\text{ ft}$
Therefore, it takes 3½ car lengths.