Ok .. here it is...

The velocity of a train is reduced from 30m/s to 10m/s over a distance of 200m. Under this constant deceleration the train travels a further distance before coming to rest. What is this distance?

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- October 29th 2006, 12:24 AMclassicstringsVectors`
Ok .. here it is...

The velocity of a train is reduced from 30m/s to 10m/s over a distance of 200m. Under this constant deceleration the train travels a further distance before coming to rest. What is this distance? - October 29th 2006, 03:02 AMclassicstrings
Posted the question now.;)

- October 29th 2006, 03:25 AMtopsquark
For the record I am assigning a positive direction in the direction of the initial velocity of the train. I am temporarily assigning an origin to be at the point where the train has a speed of 30 m/s.

To find the acceleration of the train:

[tex]a = \frac{100 - 900}{2 \cdot 200}

(The negative sign merely indicates that the acceleration is in the opposite direction to the initial velocity, meaning that it is a decleration in this case.)

Now I will reset the origin to the point where the speed is 10 m/s. We wish to find out how far the train travels from this point until it comes to rest. We may use the same equation:

-Dan - October 29th 2006, 03:34 AMclassicstrings
Hello Topsquark thanks for the answer, i'm still looking at it, but my solutions (not that I know why it is 25) says it is meant to be 25metres. Thanks!

EDIT: Ok I see your calculation error, -800 = 400a, a = -2m/s^2 - October 29th 2006, 03:41 AMclassicstrings
I get it now, thanks! It works out, after the a = -2m/s^2.

- October 29th 2006, 03:56 AMtopsquark
- October 29th 2006, 03:57 AMclassicstrings