We can solve this problem by applying kinematic equations.

a) Finding the acceleration

The kinematic equation that relates initial velocity $(v_0)$, final velocity $(v)$, acceleration $(a)$, and distance $(d)$ is:

$v^2 = v_0^2 + 2ad$

• $v_0 = 40 \, \text{m/s}$ (initial velocity),
• $v = 0 \, \text{m/s}$ (final velocity, since the train stops),
• $d = 100 \, \text{m}$ (distance).

Now, solve for $a$:

$0 = 40^2 + 2a(100)$

$0 = 1600 + 200a$

$200a = -1600$

$a = \frac{-1600}{200} = -8 \, \text{m/s}^2$

The acceleration of the train is $\boxed{-8 \, \text{m/s}^2}$.

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b) Finding the time taken

We can use another kinematic equation:

$v = v_0 + at$

• $v_0 = 40 \, \text{m/s}$,
• $v = 0 \, \text{m/s}$,
• $a = -8 \, \text{m/s}^2$.

Now, solve for $t$:

$0 = 40 + (-8)t$

$8t = 40$

$t = \frac{40}{8} = 5 \, \text{seconds}$

The time taken for the train to stop is $\boxed{5 \, \text{seconds}}$.

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Would you like more details on any of these steps? Here are five related questions:

1. What would be the distance if the initial velocity were doubled?
2. How does negative acceleration affect the motion of the train?
3. Can we calculate the force exerted by the brakes if the train's mass is known?
4. How would the time change if the acceleration were halved?
5. What other factors could influence the braking distance of a real train?

Tip: When braking or decelerating, acceleration is always negative, which means the object is slowing down.