1. ## driving problem

You are traveling at the legally posted 35 miles per hour until you reach a car in front of you that is traveling at 15 miles per hour. What distance is necessary to legally and safely pass the vehicle.

2. We need more information. Not least the between distance the '15mph car' and the '35mph' car.

3. what other information is needed?

4. Originally Posted by Mr_Green
what other information is needed?
Is acceleration constant all the time?

5. yes.

6. How far ahead is the 15 mph car ahead of the 35 mph car?

7. Originally Posted by Mr_Green
What distance is necessary to legally and safely pass the vehicle.
Explain that part to me first. What is your definition of safely and legally? This question doesn't make sense...

8. 10 feet

9. legally means that the 35 mph car must only travel at 35 mph; it cannot go above.

safely means that the 35 mph car cannot crash in the 15 mph car.

10. ok maybe let me put the question in another way

a 35 mph car is driving down the street. me 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?

the 35 mph car remains at 35 mph the whole time.

11. 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 $L$ feet,
. . the faster car $(F)$ overtakes the slower car $(S)$
. . 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 $d.$
Code:
                                L
*---------*
|    S    | →
*---------*
L
*---------*
|    F    | →
+ - - - - - -  d  - - - - - - *---------*

Relative to $S$, $F$ must travel a distance of $2L$ feet.

Relative to $S$, $F$'s speed is: . $35 - 15 \:=\:20\text{ mph} \:=\:\frac{88}{3}\text{ ft/sec}$

This will take: . $\frac{2L}{\frac{88}{3}} \:=\:\frac{3L}{44}$ seconds.

At $35\text{ mph} \:=\:\frac{154}{3}\text{ ft/sec, }F$ travels: . $\left(\frac{154}{3}\text{ ft/sec}\right) \times \left(\frac{3L}{44}\text{ sec}\right) \;=\;\frac{7}{2}L\text{ ft}$

Therefore, it takes 3½ car lengths.