217k views
1 vote
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.

1 Answer

2 votes

Answer:

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}.

User Axemasta
by
9.1k points