Let's break down the problem into three phases and calculate the velocity at each key moment.
### Phase 1: Acceleration
1. The car starts from rest (initial velocity \( v_{\text{initial}} = 0 \) m/s).
2. It accelerates at a rate of \( 3 \) m/s² for \( 4 \) seconds.
To find the velocity at the end of this phase, use the formula for velocity under constant acceleration:
[tex]\[ v = u + at \][/tex]
where:
- \( v \) is the final velocity after acceleration,
- \( u \) is the initial velocity,
- \( a \) is the acceleration,
- \( t \) is the time.
Substituting the given values:
[tex]\[ v = 0 + 3 \times 4 \][/tex]
[tex]\[ v = 12 \text{ m/s} \][/tex]
At the end of the acceleration phase, the velocity of the car is \( 12 \) m/s.
### Phase 2: Constant Velocity
1. After accelerating, the car continues at a constant velocity of \( 12 \) m/s for \( 5 \) seconds.
Since the velocity remains constant, at the end of this phase, the velocity is still \( 12 \) m/s.
### Phase 3: Deceleration
1. The car then decelerates at a rate of \( -2 \) m/s² (negative because it's deceleration) for \( 3 \) seconds.
To find the velocity at the end of this phase, use the same formula for velocity under constant acceleration:
[tex]\[ v = u + at \][/tex]
where:
- \( v \) is the final velocity after deceleration,
- \( u \) is the initial velocity (which is now \( 12 \) m/s from Phase 2),
- \( a \) is the deceleration (\( -2 \) m/s²),
- \( t \) is the time.
Substituting the given values:
[tex]\[ v = 12 + (-2) \times 3 \][/tex]
[tex]\[ v = 12 - 6 \][/tex]
[tex]\[ v = 6 \text{ m/s} \][/tex]
At the end of the deceleration phase, the final velocity of the car is \( 6 \) m/s.
### Summary
The car's final velocity after accelerating from rest, moving at a constant speed, and then decelerating is [tex]\( 6 \)[/tex] m/s.
