There are trains that operate in the Himalayas?
Let
l
be the length of a train. This particular train is on its way through the Himalayas, or at least a tunnel underneath the Himalayas. Fear not, however, because the stretch of tunnel that the train goes through is only 300 meters. It takes the train 20 seconds to completely pass in and out of the tunnel. One of the stationary ceiling lights in the tunnel is above the train for exactly 10 seconds. If the train is travels at a constant speed of
s
meters per second, find
s
+
l
.
The answer is 330.
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14 solutions
I wonder what practical value there is in determining the sum of a speed and a length
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None. I guess this is not to be practical, as it is not on "physics" field, it's in algebra... It's just to test calculation skills and problem's interpretation.
the train enters from the front and leaves when its rear is out so dont we have add train length twice???
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no man, u cant add it twice. consider the train as a point object (let our point be the train's rear). Now u start counting time as soon as the front of the train enters the tunnel. At this instance, our point object (the train's rear) is still 'L' metres away from the tunnel. We stop the time when the train's rear has emerged out of the tunnel, therefore our point object has traveled a distance of '300+L' meters.... i think that should suffice :-) :-)
If you added the length of the train twice, you would have a little of a headache trying to solve a problem like s = 3 0 0 + s
Hi this is an easy one ...isn't it
How can you add speed to length ????? O.o ??
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Well, as long as you have the units of measurement defined, you're all set. :D
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But it doesn't make sense adding speed with length !!! :( So what do u think is the unit of the answer ??
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@Dhvinay Pv – The answer doesn't have to have a unit, it's just a misdirection to the result
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@João Arruda – Hmmm... Thank you all for the assistance.. It was something strange I encountered which I can not still digest completely cos it makes no sense for me.. ;)
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@Dhvinay Pv – No problem, I guess we all have that feeling
@Dhvinay Pv – Neither of the two! If I have a car going 20 miles per hour, and a 50 meter bus, and I'm asked to find the sum of the speed (in miles per hour) and the bus (in meters), then I just say 70! It's not like I'm manipulating it in my actual equations!
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@Finn Hulse – Yes....when it is said to find s + l , we just add the magnitudes of the variables s and l and here s = 3 0 0 , l = 3 0 . So, s + l = 3 0 0 + 3 0 = 3 3 0 . I don't really see what's so confusing in this !! The units are not to be taken into consideration as just the sum of the magnitudes of the 2 variables is being asked here, nothing more !!
Actually while travelling through the tunnel the train covers the length of the tunnel as well as it own length.
In the question, the train takes 2 0 seconds to completely pass through the tunnel. Thus, speed of train is:
s = 2 0 3 0 0 + l (Equation 1 )
It takes 1 0 seconds to pass through a ceiling light which is stationary in the tunnel, which means it takes 1 0 seconds to cover its own length. Thus, speed of train is:
s = 1 0 l (Equation 2 )
Equalizing both speeds: (Equation 1 and 2 )
2 0 3 0 0 + l = 1 0 l
Cancelling the denominators by 1 0
2 3 0 0 + l = l
3 0 0 + l = 2 l
3 0 0 = 2 l − l
l = 3 0 0
Substituting the value of l to find s :
s = 2 0 3 0 0 + l or 1 0 l (You can take any of the 2 equations of find s )
s = 1 0 l
s = 1 0 3 0 0
s = 3 0
Thus, s = 3 0 and l = 3 0 0 .
Thus, the answer is: s + l = 3 0 + 3 0 0 = 3 3 0
why cant we use .............s=vt formula we get v,,,,,,,,,,,,and from it we can find l
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I see what you are saying. Actually, the formula you are referring to is just a simple rearrangement of the one that I'm using in my solution. :D
I solved the problem in the exact same way....nice problem though..easy one...:D
nice problem
does 300 meter long train passes the tunnel in 20sec???? plz figure it.
the total distance to be covered is 900 meters and speed is 30 m/sec..........isn't it what the solution determines ?
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Dear Zain, Refer to Josh Silverman's diagram above. We start measuring displacement when the Front end of the train touches the tunnel. But when the Front end of the train travels 300 meters and touches the other end of the tunnel, train is still in the tunnel. The train has to move 'l' meters more to get out of the tunnel. That means train travels 300+l meters passing through the tunnel. And you don't have to add 'l' twice.
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yap...i was wrong for my solution.... just have to focus on one end of train and the second will get itself out accordingly
(300+l)/s=20 eq1. l/s=10 eq.2 l=300, s=30
In first case the train travel distance = 300 +l in 20 seconds with speed = s. so 300 + l =20s Also, 10s = 300 Solving the two eqns we get s= 30 and l=300 so, s + l = 330
The hint is:
-The train needs 10sec to pass in 300m tunnel, s = 30m/s
-To pass in and out is 20 sec, pass out = 10sec
-To find the length of train, s×10=300
-Finally, s+l=330
question have a glitch.. it says completely pass in and out..... means it must cover 300+l+l
s.20=300+l s.10=l we get l=300,s=30 s+l=330 that's it.
'L' be length of train. 'S' be Speed (30+L)/20=S, and L/10=S
Equating L=300
thus S=30
L+S= 330
As the whole train pass through tunnel of 300 in 20 sec, and it take 10 from front of train to rear of train,
Length of train =300 m Speed of train = 300/10=30 mps
Thus s+l=330
since light bulb is above the train for 10 sec, so length of train l=10*s.........................(1)
s is speed of train,
now since train completely in and out through 300 meters long tunnel in 20 sec,therefore l+300=20*s ...............(2) put l value from (1)
we get 10*s=300 or, s=30 meter
so l=300 meter
thus l+s =330 meters
To travel 'l' meters the train takes 10 seconds To travel (l+300)meters the train takes 20 seconds i.e the train can travel 300 meters in 10seconds So speed of train = 300/10 = 30m/sec and length of train = 30 x 10 =300 so l+s = 330
speed of the train = length of the train/10 = 0.1l
speed = (300+l)/20 hence l=300 and hence s+l=330
s = (l+300)/20 ;
l = s * 10 ;
l = [(l+300)/20]*10 ;
2l = l + 300 ;
l = 300;
s = (300+300)/20 ;
ie. (l+300)/20 ;
s = 30 ;
s + l = 330
Let length of train be l and speed of train be s. speed = distance travelled / time taken
According to question, s = l/10
l/10 = [300 + l]/20
Solving, l = 300, s = 30
therefore, l + s = 330
The formula to keep in mind is r = t d , where r is rate, d is distance, and t is time. Notice in the problem that it states that it takes 20 seconds from the time that the train enters, until it completely leaves, not just for it to go the 300 meters. We can set up the following equation:
s = 2 0 3 0 0 + l
As for the other information given, it takes 10 seconds for the train traveling at speed s to go length l . Thus:
s = 1 0 l
Rearranging gives
l = 1 0 s
Substituting into the first equation produces
s = 2 0 3 0 0 + 1 0 s
Solving, we find that s = 3 0 and l = 3 0 0 . Their sum is 3 3 0 .