The speed of the blue train - Crafts & Hobbies

No one sells spare tracks. It comes with 20 feet of track. If you go to a hobby shop that tells train sets and modelmaking supplies, you might be lucky if it is a standard gauge and connection.

https://www.google.com/search?q=Blue+Hat+Toy+Company+replacement+train+track

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Hi, Jamie and the usual suspects are:
1. Excessive choke use.
2. High-speed riding style.
3. Idle speed improperly adjusted.
4. Dirty air cleaner element (may also affect idle speed).
5. Restricted fuel tank vent system.
6. Vacuum piston assembly malfunction.
7. Excessive accelerator pump output.
8. Plugged or restricted bowl vent.
9. Float level too high.
10. Loose carburetor jets.
11. Worn or damaged needle or needle jet.
12. Choke valve not seated or leaking.
13. Plugged air jets or passages.
14. Low air pressure in the tires.
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You use the train controller. Set the control to zero for off and then rotate it through the scale up to one hundred for full speed. Use the sliding direction button above the speed control to select direction of the train. As far as I know there is no on/off switch on the loco itself.

Sounds like the "variable resistor" that allows for a variable speed is either defective, or has become corroded on it's switch contacts. You might get lucky with some electrical contact cleaner spray and spraying it in or around the variable speed switch. Otherwise, I'd just invest in a new transformer.

If the train's still going the right way, up about ((5 *60)/25) km/h. Classical relativity doesn't change it much at that person's speed, since c >> 25kmph, but that's not to say quantum relativity will continue to favor the Standard Model with the information given for meeting in your OS.

You have two variables on the question. That is the Speed of the Train and the actual Time taken.
So you shouold be have two equations to resolve the two variables.
Lets assume the speed to be x and the actual time in minutes to be y.
So the 1st equation,
If the train goes with the actual or the normal speed x the train reaches within the time y.
we know that speed = distance X Time
so X = Distance x Y or
Distance = X/Y ------- 1
Next the 2nd equation.
[{(12.5/100) x X} + X] = Distance x (y-20) or
Distnace = [{(12.5/100) x X} + X] / (y-20) ------ 2
Now you can equate 1 and 2 to get the results.

The stopping distance is proportional to square of speed.

that is 90ft proportional to 40^2 mph.

We can write as 90 = X * 40^2. ( X is a constant of proportionality) .

Which implies X = 90 / 1600 = .05625.

Assume S be the distance at which car stops when the speed is 90 mph.

S = X * 90^2. ( since X does not vary as same car is used)

=> S =.05625 * 8100 = 455.625 ft.

Thank you.

Let X be the time taken when train is moving in its normal speed.

Y be the normal speed of the train.

Speed = distance/time

that is X = 400 / Y

or Y = 400 / X --------- (1)

When speed is decreased by 5 km/hr time increases by 4 hour

that is X + 4 = 400 / Y -5

= X + 4 = 400 X / 400 - 5X ( substitute Y = 400 /X we get this equation)


on solving we get 5X^2 + 20x - 1600 = 0

we get two solutions X = 16 and -20

-20 is neglected (time cannot be negative)

substitute X in equation (1)

Y = 400 / 16 = 25 km/hr

So normal speed of train is 25 km/hr.