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The Car and the Train (Posted on 2008-01-30) Difficulty: 3 of 5
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

See The Solution Submitted by Brian Smith    
Rating: 2.0000 (1 votes)

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Solution solution | Comment 1 of 2
  • Let s be the speed of the train
  • Let a be the car's acceleration
  • Let L be the length of the train
  • Let t1 be the interval between the front of the train passing the infinitely small car and the back of the train passing the car.
  • Let d1 be the distance traveled by the car during time t1
  • Let t2 be the interval between the back of the train passing the car and the car catching up to the front of the train.
  • Let d2 be the distance traveled by the car during time t2
    During time t1, the train travels s*t1 units, but the car needs to have traveled L units less than this, so:

s*t1 - L = a*(t1)^2 / 2 = d1  (eq. 1)

Since the car is traveling at the same speed as the train when it is even with the back of the train:

a*t1 = s    (eq. 2)

Therefore

s*t1 - L = s*t1/2   (eq. 3)

or

L = s*t1/2 = a*(t1)^2 / 2 = d1  (eq. 4)

When the car is even with the back of the train, it is at speed zero relative to the train (they are going at the same speed). Considered from the frame of reference of the train, then:

a*(t2)^2 / 2 = L   (eq.5)

but previously,

a*(t1)^2 / 2 = L   (from eq. 4)

so t2=t1.    (eq. 6)

During the interval t2, the car has moved L units relative to the train, and the train has moved s*t2 units, which is the same as s*t1. From a previous equation (eq. 4), this is 2*L.

So the car has in all moved L + 2*L = 3*L units, and so has the train, since they start and finish with the car even with the front of the train.

When the car reaches the front of the train and its speed is the speed limit, it is going at a speed of a*2*t1.  Since a*t1*t1 = 2*L, a*2*t1 = 4*L/t1.

But L/t1 = s/2 (from eq. 4), so the speed limit is 4*s/2 = 2*s.

As a check,

s = 27.5 mph
L = 1/4 mile (1320 feet or 440 yards)
t1 = 1/55 hour (60/55 min or 1 min, 5 5/11 sec)
a = 55*27.5 = 1512.5 mph/hr = 25.2083333... mph/min)
              front of train   car     back of train car speed
initial time        0           0       -0.25 mi         0
1/55 hour        0.5 mi       0.25 mi    0.25 mi       27.5 mph
2/55 hour        1.0 mi       1.00 mi    0.75 mi       55.0 mph

Edited on January 30, 2008, 1:15 pm
  Posted by Charlie on 2008-01-30 12:46:29

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