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?
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?
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