Two cars left place A simultanecously and reached the place B in 2 hrs. The first car travelled half the distance with a speed of 30 km/hr and the other half at a speed of 45 km/hr. The second car at the same time covered the entire distance with a constant acceleration starting from rest . Find the acceleration of the second car.

Options

A. `36 km//hr^(2)`
B. `18 km//hr^(2)`
C. `72 km//hr^(2)`
D. `144 km//hr^(2)`

Answer

To solve the problem, we need to find the acceleration of the second car that travels the same distance as the first car but does so with constant acceleration starting from rest. Let's break down the solution step by step.

### Step 1: Determine the distance traveled by the first car
Let the total distance from A to B be \( L \). The first car travels half the distance (\( \frac{L}{2} \)) at a speed of 30 km/hr and the other half (\( \frac{L}{2} \)) at a speed of 45 km/hr.

### Step 2: Calculate the time taken for each half of the journey
1. **Time for the first half (A to C)**:
\[ T_1 = \frac{\text{Distance}}{\text{Speed}} = \frac{\frac{L}{2}}{30} = \frac{L}{60} \text{ hours} \]

2. **Time for the second half (C to B)**:
\[ T_2 = \frac{\text{Distance}}{\text{Speed}} = \frac{\frac{L}{2}}{45} = \frac{L}{90} \text{ hours} \]

### Step 3: Relate the times \( T_1 \) and \( T_2 \)
The total time for the journey is given as 2 hours:
\[ T_1 + T_2 = 2 \]
Substituting the expressions for \( T_1 \) and \( T_2 \):
\[ \frac{L}{60} + \frac{L}{90} = 2 \]

### Step 4: Find a common denominator and solve for \( L \)
The common denominator for 60 and 90 is 180:
\[ \frac{3L}{180} + \frac{2L}{180} = 2 \]
\[ \frac{5L}{180} = 2 \]
Multiplying both sides by 180:
\[ 5L = 360 \implies L = \frac{360}{5} = 72 \text{ km} \]

### Step 5: Apply the equations of motion for the second car
The second car travels the entire distance \( L \) with constant acceleration starting from rest. The equation of motion is:
\[ S = Ut + \frac{1}{2}at^2 \]
Since the initial speed \( U = 0 \):
\[ L = \frac{1}{2} a t^2 \]
Substituting \( L = 72 \) km and \( t = 2 \) hours (which is \( 2 \times 3600 = 7200 \) seconds):
\[ 72 = \frac{1}{2} a (2^2) \]
\[ 72 = \frac{1}{2} a (4) \]
\[ 72 = 2a \implies a = \frac{72}{2} = 36 \text{ km/hr}^2 \]

### Final Answer
The acceleration of the second car is \( 36 \text{ km/hr}^2 \).

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