The problem appears to be in Indonesian, and here's a translated version:

4. A train accelerates at 2 m/s² for some time from rest, then decelerates with a deceleration of 4 m/s² until it comes to a stop. The total distance covered by the train is:

A. 50 m
B. 100 m
C. 150 m
D. 200 m
E. 250 m

Let's solve this step by step:

1. Acceleration Phase:

   • Initial velocity, $u = 0$ (starting from rest)
   • Acceleration, $a = 2 \text{ m/s}^2$
   • Let $t$ be the time during which the train accelerates

   The velocity after time $t$ is given by: $v = u + at = 0 + 2t = 2t \text{ m/s}$

   The distance traveled during acceleration is: $s_1 = ut + \frac{1}{2}at^2 = 0 + \frac{1}{2} \cdot 2 \cdot t^2 = t^2 \text{ meters}$

2. Deceleration Phase:

   • Initial velocity, $v = 2t \text{ m/s}$
   • Deceleration, $a = -4 \text{ m/s}^2$ (negative because it's decelerating)
   • Final velocity, $v_f = 0$ (the train comes to a stop)

   Using the equation $v_f^2 = v^2 + 2as$, we find the distance during deceleration: $0 = (2t)^2 + 2(-4)s_2$ $4t^2 = 8s_2$ $s_2 = \frac{4t^2}{8} = \frac{t^2}{2}$

3. Total Distance: $s_{\text{total}} = s_1 + s_2 = t^2 + \frac{t^2}{2} = \frac{3t^2}{2}$

To find the total distance $s_{\text{total}}$, we need to use the choices to match $\frac{3t^2}{2}$ to a given answer. If $t = 10$, then: $\frac{3 \cdot 10^2}{2} = \frac{300}{2} = 150 \text{ m}$

Thus, the answer is C. 150 m.

Would you like more details or have questions?

Related Questions:

1. What if the acceleration was different, how would it change the total distance?
2. How does the deceleration affect the total time taken for the journey?
3. What would be the result if the initial speed was not zero?
4. How do we calculate the distance if the acceleration and deceleration were different in magnitude?
5. Why does the equation for distance use acceleration squared?

Tip:

Remember to keep track of signs in acceleration problems, as positive and negative values represent different directions (acceleration vs. deceleration).