13. Two trains A and B running on parallel lines enter a tunnel 2 lon
long at the opposite ends at the same time. They meet in 1 minutes and
A has then travelled in the tunnel 400 m more than B. In 4 seconds they
cross each other, and 1 minute later A has completely passed through the
tunnel. Find the lengths and the rates of the trains.
Answers
Step-by-step explanation:
Two trains X and Y enter at the same time a tunnel 1 Km long at the opposite ends.
They meet in 45 seconds, after X has traveled 200 meters more than Y.
:
Find the distance each train has traveled when they meet
let a = distance traveled by x when they meet
then
(a-200) = distance traveled by y when they meet
therefore
a + (a-200) = 1000 meters
2a = 1000 + 200
2a = 1200
a = 600 meters traveled by x, and 400 meters traveled by y
then
600/45 = 13.33 m/sec is the speed of x
and
400/45 = 8.89 m/sec is the speed of y
:
They cross each other in 4.5 seconds
Find the total Length of the two trains
L = 4.5(13.33+8.89)
L = 100 meters
:
and 30 seconds later X is completely out of the tunnel.
total time for x to enter and clear the tunnel
45 + 4.5 + 30 = 79.5 seconds
Find how far the train X travels in 79.5 sec at 13.33 m/sec
79.5*13.333 = 1060 meters
the tunnel is 1000 meters, therefore the train is 60 meters long
:
Find the length of y
100 - 60 = 40 meters is the length of train y
:
find the length and speed of each of the trains.
X is 60 meters long
Y = 40 meters long
and
X = %2813.333%2A3600%29%2F1000 = 48 km/hr
Y = %288.889%2A3600%29%2F1000 = 32 km/hr
:
:
2) A train after traveling 50 Km. Meets with an accident and then proceeds at ¾ th of its former speed and arrives at its destination 35 minutes late.
let s = train's normal speed
then
.75s = train's speed after the the accident
Let d = distance of the whole trip
then
(d-50) = distance at a speed of .75s
Write a time equation
%28d-50%29%2F%28.75s%29 - %28d-50%29%2Fs = 35%2F60
multiply by 60s so get rid of the denominators, resulting in:
80(d-50) - 60(d-50) = 35s
80d - 4000 - 60d + 3000 = 35s
20d - 35s - 1000 = 0
20d - 35s = 1000
then we are dealing with a distance of (d-74)
Had the accident happened 24 Km. Further on, it would have reached the destination only 23 minutes late.
%28d-74%29%2F%28.75s%29 - %28d-74%29%2Fs = 23%2F60
multiply by 60s so get rid of the denominators, resulting in:
80(d-74) - 60(d-74) = 23s
80d - 5920 - 60d + 4440 = 23s
20d - 23s - 1480 = 0
20d - 23s = 1480
:
Use elimination here
20d - 23s = 1480
20d - 35s = 1000
------------------subtraction eliminates d, find s
12s = 480
s = 480/12
s = 40 km/hr is the normal speed (30 km/hr is the slower speed)
:
Find the speed of the train and the distance.
20d - 23(40) = 1480
20d - 920 = 1480
20d = 1480 + 920
20d = 2400
d = 2400/20
d = 120 km
:
:
Check that in the other equation
20(120) - 35(40) =
2400 - 1400 = 1000; confirms our solution of:
speed: 40 km/hr, normal speed
dist: 120 km
Step-by-step explanation:
The relative speed of B wrt A will be V
BA
=V
B
−V
A
=20−20=0
their relative acceleration will be a
BA
=a
B
−a
A
=1−0=1m/s
2
the distance traveled will be in s=ut+
2
1
at
2
=0+
2
1
1×(50)
2
=1250meter
if initial separation was x then x+400m=s or x=1250−400=850m
because now the drivers are side by side of each other so the length of train A was added.