An automobile starts from rest and accelerates to a final velocity in two stages along a straight road. Each stage occupies the same amount of time. In stage 1, the magnitude of the car's acceleration is `3.0 m//s^(2)`. The magnitude of the car's velocity at the end of stage 2 is 2.5 times greater than it is at the end of stage 1. Find the magnitude of the acceleration is stage 2.

Options

A. `4.5 m//s^(2)`
B. `3.0 m//s^(2)`
C. `3.8 m//s^(2)`
D. `2.2 m//s^(2)`

Answer

To solve the problem step by step, we will follow the motion equations and the relationships given in the question.

### Step 1: Understand the problem
The automobile starts from rest and accelerates in two stages. We know:
- Stage 1 acceleration \( a_1 = 3.0 \, \text{m/s}^2 \)
- The final velocity at the end of stage 2 is \( 2.5 \) times the velocity at the end of stage 1.
- The time taken for both stages is the same.

### Step 2: Define variables
Let:
- \( t \) = time for each stage
- \( v_b \) = velocity at the end of stage 1
- \( v_c \) = velocity at the end of stage 2
- \( a_2 \) = acceleration during stage 2 (which we need to find)

### Step 3: Calculate the velocity at the end of stage 1
Using the first equation of motion for stage 1:
\[ v_b = u + a_1 t \]
Since the car starts from rest, \( u = 0 \):
\[ v_b = 0 + 3.0 \cdot t = 3.0t \]

### Step 4: Relate the velocities
According to the problem, the velocity at the end of stage 2 is:
\[ v_c = 2.5 v_b \]
Substituting the expression for \( v_b \):
\[ v_c = 2.5 \cdot (3.0t) = 7.5t \]

### Step 5: Calculate the acceleration during stage 2
Using the first equation of motion for stage 2:
\[ v_c = v_b + a_2 t \]
Substituting the known values:
\[ 7.5t = 3.0t + a_2 t \]

### Step 6: Solve for \( a_2 \)
Rearranging the equation gives:
\[ 7.5t - 3.0t = a_2 t \]
\[ 4.5t = a_2 t \]
Dividing both sides by \( t \) (assuming \( t \neq 0 \)):
\[ a_2 = 4.5 \, \text{m/s}^2 \]

### Final Answer
The magnitude of the acceleration in stage 2 is \( 4.5 \, \text{m/s}^2 \).

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