Question

A train is traveling at a speed of 80 km/h when the conductor applies the brakes. The train slows with a constant acceleration of magnitude .5 m/s^2. We want to find the distance the train travels from the time the brakes are applied until the train comes to a complete stop. Which kinematic formula would be most useful to solve for the target unknown?

Answer 1

Answer:

The train will travel 493.817 meters until it stops completely.

Step-by-step explanation:

In this case, we know that train applies the brakes and decelerates at constant rate until rest is reached after travelling an unknown distance. Travelled distance ($\Delta s$), measured in meters, can be found by using this kinematic formula:

$v^{2} = v_{o}^{2}+2\cdot a \cdot \Delta s$ (Eq. 1)

Where:

$v_{o}$ - Initial speed, measured in meters per second.

$v$ - Final speed, measured in meters per second.

$a$ - Acceleration, measured in meters per square second.

Now travelled distance is cleared within the formula:

$\Delta s = \frac{v^{2}-v_{o}^{2}}{2\cdot a}$

If we know that $v_{o} = 22.222\,\frac{m}{s}$, $v = 0\,\frac{m}{s}$ and $a = -0.5\,\frac{m}{s^{2}}$, then the distance travelled by the train is:

$\Delta s = \frac{\left(0\,\frac{m}{s} \right)^{2}-\left(22.222\,\frac{m}{s} \right)^{2}}{2\cdot \left(-0.5\,\frac{m}{s^{2}}\right) }$

$\Delta s = 493.817\,m$

The train will travel 493.817 meters until it stops completely.