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The problem describes a vehicle that starts from rest, accelerates at \(1.5 \, \text{m/s}^2\), and travels a total distance of 120 m, including a portion where it moves at constant speed before decelerating at \(2 \, \text{m/s}^2\) to stop.

Here’s how we can break down the problem:

### Part (a) - Finding velocity at 4s, 10s, and 15s

#### Phase 1: Acceleration
- Initial velocity (\(v_0\)) = 0 m/s (starts from rest).
- Acceleration (\(a\)) = \(1.5 \, \text{m/s}^2\).
- The first 48 meters are traveled under acceleration.

Using the equation of motion:
\[
v^2 = v_0^2 + 2a \cdot d
\]
Substitute \(v_0 = 0\), \(a = 1.5 \, \text{m/s}^2\), and \(d = 48 \, \text{m}\), to find the final velocity after 48 meters.

#### Phase 2: Constant Velocity
- The vehicle maintains a constant speed after reaching 48 m until it starts decelerating.
- We'll need to determine how long the vehicle stays in this phase.

#### Phase 3: Deceleration
- The vehicle decelerates at \(2 \, \text{m/s}^2\) until it stops.
  
### Part (b) - Finding the distance traveled at constant velocity
- The total distance covered by the vehicle is 120 m, and we already know 48 m is traveled during acceleration.
- The remaining distance will be split between the constant velocity phase and the deceleration phase.

Now, I'll calculate the velocities at the specified times and the distance traveled at constant speed.The velocity of the vehicle after the acceleration phase, having traveled 48 meters, is \(12 \, \text{m/s}\).

Next, I will calculate the velocity at 4s, 10s, and 15s. Then, I will determine the distance traveled during the constant velocity phase.At 4 seconds, the velocity of the vehicle is \(6 \, \text{m/s}\), and at 10 seconds, the velocity is \(15 \, \text{m/s}\). 

Since the vehicle reaches \(12 \, \text{m/s}\) after 48 meters, it means at 10 seconds, the vehicle would have completed the acceleration phase.

Now, let's calculate the total distance traveled during the constant velocity phase.The vehicle travels 36 meters during the constant velocity phase and 36 meters during the deceleration phase.

### Summary:
- The velocity of the vehicle at 4 seconds is \(6 \, \text{m/s}\).
- The velocity at 10 seconds is \(15 \, \text{m/s}\) (before the acceleration ends).
- The vehicle travels 36 meters at constant speed before decelerating.

Would you like any additional details or clarification? 

Here are 5 related questions for further exploration:
1. How do you calculate the time spent in each phase of motion?
2. What is the relationship between acceleration and distance in uniformly accelerated motion?
3. How does the deceleration rate affect the stopping distance?
4. Can you derive the total time for the trip based on the given accelerations and distances?
5. How would the final velocity change if the acceleration were doubled?

**Tip:** Always break down motion problems into distinct phases (acceleration, constant speed, and deceleration) to simplify the calculations.

Un vehículo parte del reposo y acelera a razón a 1.5 m/s², mientras recorre una distancia de 48 m, luego mantiene su velocidad constante durante cierto tiempo y posteriormente frena a razón de 2 m/s² hasta detenerse. Si recorre una distancia total de 120 m. Calcular: a) La velocidad a los 4s, 10s y 15s. b) El espacio recorrido con velocidad constante.

Solve a physics problem where a vehicle accelerates at 1.5 m/s² over 48 meters, maintains constant speed, and decelerates at 2 m/s² to stop. Learn how to calculate velocities at 4s, 10s, and 15s, and find the distance traveled at constant speed.

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