Train Problems - Speed and Distance

If you understand the following key concepts then questions on speed and distance, especially problems on train are easy to tackle.

The Basic formula for Speed and Distance holds true for train problems too.

D = V * T
D = Distance
V = Speed
T = Time

To calculate speed of a train relative to an object we use the formula:

Relative Speed (V’) = V (t) – V (o) where V (t) is the speed of the train and V (o) is the speed of the object.

How about distance covered by the train relative to the object?

D = L (Train) + L (Object)

Based on the length of the object there are four cases

Case 1:  Train crosses a stationary object, which has no length

L (Train) = V (t) * T

Case 2:  Train crosses a stationary object with length L (Object)

L (Train) + L (Object) = V (t) *T

Case 3: Train crosses a moving object, which has no length

Same Direction: L (Train) = (V (t) – V (o)) * (T)
Opposite Direction: L (Train) = (V (t) + V (o)) * (T)

Case 4: Train crosses a moving object with length L (Object)

Same Direction: L (Train) + L (Object) = (V (t) – V (o)) * T
Opposite Direction: L (Train) + L (Object) = (V (t) + V (o)) * T

Let’s apply what we have learned

A train travelling at a speed of 75 mph enters a tunnel 3 1/2 miles long. The
train is 1/4 mile long. How long does it take the train to pass through the
tunnel from the moment the front enters to the moment the rear emerges?

a) 2.5 min
b) 3 min
c) 3.2 min
d) 3.5 min

Don’t get confused by the last sentence (“train to pass through the
tunnel from the moment the front enters to the moment the rear emerges”)

Here the stationary object is the tunnel. The question is how long it takes the train to pass the stationary object with length L (Object).

V (T) = 75 mph
L (Object) = 3 1/2 miles
L (Train) = ¼ miles

L (Object) + L (Train) = V (T) * T
15/4 = 75 * T
T = (1/20) hr = 3 minutes

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