A train is moving at a constant speed on a track parallel to a road. As the front of the train passes a stopped car, the car starts accelerating at a constant rate. When the back of the train catches up to the car, the car is moving as fast as the train. As the car continues to accelerate, it catches up to the front of the train right when the car reaches the speed limit.

Express the distance traveled in terms of the length of the train.
Express the speed limit in terms of the speed of the train.

Formulas:
distance = speed * time
distance = (1/2) * acceleration * time^2
speed = time * acceleration

Assuming that the distance desired is the distance travelled by the car (or the front of the train):

 

Let L be the length of the train

Let V be the speed of the train

Let a be the acceleration of the car

Let t1 be the time when the car reaches the back of the train and matches the train's speed

Let t2 be the time when the car catches up to the front of the train

 

At time = t1:

The car has traveled the same distance as the train, less the length of the train:

.5 * a * t1^2 + L = V * t1    EQ 1

& the car's speed equals the train's' speed

a * t1 = V    EQ 2

 

Solving EQ 2 for t1 and substituting in EQ1, and solving for "a":

 

a = V^2 / (2*L)  EQ 3

 

Moving to t2, the distance travelled by the car and train are equal (car is again at the front of the train where it started):

 

.5 * a * t2^2 = V * t2  EQ 4

 

Substituting EQ 3 into EQ 4 and rearranging a bit:

 

(1/4)(1/L)(V^2* t2 ^2) = V * t2

 Remembering that we want to solve for the distance travelled, which is equal to V * t2, substitute D=V * t2:

 

(1/4)(1/L)(D^2)=D  or  D = 4 * L    ß answer to part 1

 

For part 2, we only need to realize that the cars final speed=speed limit is a * t2, and solve EQ4 for that quantity:

 

.5 (speed limit) * t2 = V * t2 or speed limit = 2* V

 

Summarizing

 

The speed limit is 2 times the speed of the train

The distance travelled is 4 times the length of the train

Edited on February 3, 2008, 5:12 pm

Posted by Kenny M on 2008-02-01 00:13:30