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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- Oct 28th 2006, 11:24 PMclassicstringsVectors`
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? - Oct 29th 2006, 02:02 AMclassicstrings
Posted the question now.;)

- Oct 29th 2006, 02:25 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:

$\displaystyle v^2 = v_0^2 + 2a(x - x_0)$

$\displaystyle 10^2 = 30^2 + 2 \cdot a \cdot 200$

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

$\displaystyle a = -2 \, m/s^2$

(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:

$\displaystyle v^2 = v_0^2 + 2a(x - x_0)$

$\displaystyle 0^2 = 10^2 + 2(-2)(x - 0)$

$\displaystyle x = \frac{0 - 100}{2 \cdot -4}$

$\displaystyle x = 25 \, m$

-Dan - Oct 29th 2006, 02:34 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 - Oct 29th 2006, 02:41 AMclassicstrings
I get it now, thanks! It works out, after the a = -2m/s^2.

- Oct 29th 2006, 02:56 AMtopsquark
- Oct 29th 2006, 02:57 AMclassicstrings