Question

A train is traveling at a speed of 60 km/hr when the conductor applies the brakes. The train slows with a constant acceleration of magnitude 0.5 m/s2. To find the distance the train travels from the time the conductor applies the brakes until the train comes to a complete stop.

Answer 1

Approximately
`278\; {\rm m}`.

Step-by-step explanation:

The following SUVAT equation relates displacement
`x` to initial velocity
`u`, final velocity
`v`, and acceleration
`a`:

`\displaystyle x = (v^(2) - u^(2))/(2\, a)`.

In this question:

• Final velocity:
  `v = 0\; {\rm m\cdot s^(-1)}` since the train has completed stopped.
• Initial velocity:
  `u = 60\; {\rm km\cdot h^(-1)}`, which needs to be converted into the standard unit of meters per second.
• Acceleration:
  `a = (-0.5)\; {\rm m\cdot s^(-2)}`, which is negative since the train is slowing down.

Apply unit conversion and ensure that all quantities are measured in standard units:

`\begin{aligned} u &= (60\; {\rm km\cdot \text{hour}^(-1)}) * \frac{1000\; {\rm m}}{1\; {\rm km}} * \frac{1\; {\text{hour}}}{3600\; {\rm s}} \\ &= (50)/(3)\; {\rm m\cdot s^(-1)}\end{aligned}`.

Substitute the value of
`v`,
`u`, and
`a` into the expression for
`x` to obtain:

`\begin{aligned}x &= (v^(2) - u^(2))/(2\, a) \\ &= (0^(2) - (50 / 3)^(2))/(2\, (-0.5))\; {\rm m} \\ &\approx 278\; {\rm m}\end{aligned}`.