Velocity of train if it passes 2km bridge in 110sec and a point in 10s

In summary: So 20 km in 1000 seconds, or 72 km/hr.In summary, a train passes through an electric post in 10 seconds and through a 2km long bridge in 110 seconds with the same velocity. The options for the train's velocity are 18kmph, 36kmph, 72kmph, and 90kmph. After calculating the change in displacement of the train with respect to the bridge, it was determined that the train's velocity is 72kmph. The time from when the train's engine entered the bridge until the train completely left the bridge is 110 seconds. The displacement was calculated with respect to the start of the bridge in the first diagram and with respect to the end
  • #1
Govind_Balaji
83
0

Homework Statement


A train passes through an electric post in 10 seconds. With same velocity, it passes through a 2km long bridge in 110 seconds.Find the velocity of the train:

a)18kmph b)36kmph c)72kmph d)90kmph.


Homework Equations







The Attempt at a Solution



The train passes through the electric post in 10s. Let it velocity be ##v##.

##\Rightarrow \text{Length of the train} = 10v##

Then I calculated the change in displacement of the train with respect to the bridge when it passes through.

https://www.physicsforums.com/attachment.php?attachmentid=70093&stc=1&d=1401201462

##\text{Displacement of train from bridge after overtake}=2km+10v##
##\text{Displacement of train from bridge after overtake}=-2km##

##\delta s=(2km+10v)-(-2km)=4km+10v##

It took 110 seconds for the train.
Its displacement=##110v##

##\Rightarrow Displacement=110v=4km+10v##

##v=\frac{4*3600}{100}km/h \Rightarrow v=108km/h##

But itsn't in the options.
 
Last edited:
Physics news on Phys.org
  • #2
Govind_Balaji said:
The train passes through the electric post in 10s. Let it velocity be ##v##.

##\Rightarrow \text{Length of the train} = 10v##

Then I calculated the change in displacement of the train with respect to the bridge when it passes through.

##\text{Displacement of train from bridge after overtake}=2km+10v##
The train covers this 2000 + 10v distance in 110 secs, so speed v = ...
 
  • #3
Whah? How did you draw that tick? ##\text{Nascent}O_2##?
 
  • #4
NascentOxygen said:
The train covers this 2000 + 10v distance in 110 secs, so speed v = ...

If i take it 2km+10v, then I get 72km/h.

But why should I take 2km+10v instead of 4km+10v?

Isn't 110 seconds is the time from when the train's engine entered the bridge till the train completely left the bridge?

Or is 110 seconds is the time from when the train's engine entered the bridge till the train's engine left the bridge?

Also, how did you place that tick??
 
  • #5
adjacent said:
Whah? How did you draw that tick? ##\text{Nascent}O_2##?

Code:
color="seagreen"
size="7"
✔
/size
/color

Just use the brackets. Result:


I do not know if there is another way.
 
  • #6
Govind_Balaji said:
If i take it 2km+10v, then I get 72km/h.

But why should I take 2km+10v instead of 4km+10v?
Draw a diagram showing the last carriage exiting the bridge. Where is the engine at this juncture?
 
  • #7
Govind_Balaji said:
Isn't 110 seconds is the time from when the train's engine entered the bridge till the train completely left the bridge?
It's clear from the figure, that the above interpretation is the correct one.

(There's also a quite simple way to do this problem, but I suspect that you're expected to do it with Algebra.)
Or is 110 seconds is the time from when the train's engine entered the bridge till the train's engine left the bridge?
 
  • #8
NascentOxygen said:
Draw a diagram showing the last carriage exiting the bridge. Where is the engine at this juncture?

I made a small mistake. I calculated displacement in first diagram w.r.t. the start of the bridge.

But I calculated the displacement in second diagram w.r.t the end of the bridge.

Thanks for all of your help.
 
  • #9
Govind_Balaji said:
I made a small mistake. I calculated displacement in first diagram w.r.t. the start of the bridge.

But I calculated the displacement in second diagram w.r.t the end of the bridge.

Thanks for all of your help.
As a check:

From the time the locomotive (assumed at the front of the train) gets to the end of bridge, to the time the last train car just clears the end of the bridge, 10 seconds elapse.

So it took the locomotive 100 seconds to cross the bridge.

That's 2km in 100 seconds.
 

FAQ: Velocity of train if it passes 2km bridge in 110sec and a point in 10s

1) How do you calculate the velocity of a train passing a 2km bridge in 110 seconds?

The velocity of an object is calculated by dividing the distance traveled by the time it took to travel that distance. In this case, the distance is 2km and the time is 110 seconds. Therefore, the velocity of the train passing the bridge would be 2km/110s = 0.0182 km/s.

2) Can you explain the significance of the point in 10 seconds when calculating the velocity of the train?

The point in 10 seconds is used as a reference point to calculate the velocity of the train. It is the point where the train is passing by at a specific time, which is 10 seconds. By measuring the distance the train has traveled in that time, we can determine its velocity.

3) What units should be used when expressing the velocity of the train?

The velocity of an object is typically expressed in units of distance per time, such as meters per second or kilometers per hour. In this case, the velocity of the train would be expressed in kilometers per second (km/s).

4) Is there any other information needed to calculate the velocity of the train?

In order to accurately calculate the velocity of the train, we would also need to know the distance from the point to the end of the bridge. This would help us determine the total distance traveled by the train and therefore its velocity.

5) How does the velocity of the train passing the bridge compare to the velocity at the point in 10 seconds?

Since the train is traveling at a constant velocity, the velocity at the point in 10 seconds should be the same as the velocity when passing the 2km bridge. This is because the train covers the same distance in both cases, just at different times. Therefore, the velocity at the point in 10 seconds would also be 0.0182 km/s.

Back
Top