Two motor cars start from A simultaneously & reach B after 2 hour. The first car travelled half the distance at a speed of `v_(1)=30 km hr^(-1)` & the other half at a speed of `v_(2)=60 km hr^(-1)`. The second car covered the entire with a constant acceleration. At what instant of time, were the speeds of both the vehicles same? Will one of them overtake the other enroute?

Explanation

To solve the problem, we need to analyze the motion of both motor cars and determine when their speeds are the same and whether one overtakes the other.

### Step 1: Determine the total distance traveled by both cars.
Both cars start from point A and reach point B in 2 hours. Let's denote the total distance from A to B as \( S \).

### Step 2: Calculate the distance traveled by the first car.
The first car travels half the distance at a speed of \( v_1 = 30 \, \text{km/hr} \) and the other half at \( v_2 = 60 \, \text{km/hr} \).

1. For the first half of the distance:
\[ S_1 = \frac{S}{2} = v_1 \cdot T_1 \Rightarrow \frac{S}{2} = 30 \cdot T_1 \]
Thus,
\[ S = 60 \cdot T_1 \quad \text{(1)} \]

2. For the second half of the distance:
\[ S_2 = \frac{S}{2} = v_2 \cdot T_2 \Rightarrow \frac{S}{2} = 60 \cdot T_2 \]
Thus,
\[ S = 120 \cdot T_2 \quad \text{(2)} \]

### Step 3: Relate \( T_1 \) and \( T_2 \).
Since the total time is 2 hours:
\[ T_1 + T_2 = 2 \quad \Rightarrow \quad T_2 = 2 - T_1 \quad \text{(3)} \]

### Step 4: Substitute \( T_2 \) in equation (2).
Substituting equation (3) into equation (2):
\[ S = 120 \cdot (2 - T_1) \]
Equating equations (1) and (2):
\[ 60 \cdot T_1 = 120 \cdot (2 - T_1) \]
Expanding:
\[ 60T_1 = 240 - 120T_1 \]
Combining like terms:
\[ 180T_1 = 240 \quad \Rightarrow \quad T_1 = \frac{240}{180} = \frac{4}{3} \, \text{hours} \]

### Step 5: Calculate \( S \).
Using \( T_1 \) in equation (1):
\[ S = 60 \cdot \frac{4}{3} = 80 \, \text{km} \]

### Step 6: Calculate \( T_2 \).
Using \( T_1 \) in equation (3):
\[ T_2 = 2 - \frac{4}{3} = \frac{2}{3} \, \text{hours} \]

### Step 7: Calculate the acceleration of the second car.
The second car travels the entire distance \( S = 80 \, \text{km} \) with constant acceleration. Using the equation of motion:
\[ S = ut + \frac{1}{2} a t^2 \]
Where \( u = 0 \) (initial speed) and \( t = 2 \, \text{hours} = \frac{2 \times 3600}{1000} = 7.2 \, \text{s} \):
\[ 80 = 0 + \frac{1}{2} a (2)^2 \]
Thus,
\[ 80 = 2a \quad \Rightarrow \quad a = 40 \, \text{km/hr}^2 \]

### Step 8: Determine when the speeds of both vehicles are the same.
The speed of the second car after time \( t \) is given by:
\[ v = u + at = 0 + 40t \]
Setting this equal to the speeds of the first car:
1. For \( 30 \, \text{km/hr} \):
\[ 40t = 30 \quad \Rightarrow \quad t = \frac{30}{40} = \frac{3}{4} \, \text{hours} \]
2. For \( 60 \, \text{km/hr} \):
\[ 40t = 60 \quad \Rightarrow \quad t = \frac{60}{40} = 1.5 \, \text{hours} \]

### Step 9: Determine if one car overtakes the other.
To check if the second car overtakes the first, we can compare their distances at various times.

1. At \( t = \frac{4}{3} \) hours:
- Distance of Car 1:
- First half: \( 30 \cdot \frac{4}{3} = 40 \, \text{km} \)
- Distance of Car 2:
- \( S = ut + \frac{1}{2} a t^2 = 0 + \frac{1}{2} \cdot 40 \cdot \left(\frac{4}{3}\right)^2 = \frac{1}{2} \cdot 40 \cdot \frac{16}{9} = \frac{320}{9} \approx 35.56 \, \text{km} \)

Since the distance of Car 1 is greater than that of Car 2 at \( t = \frac{4}{3} \) hours, Car 2 does not overtake Car 1.

### Conclusion
1. The speeds of both vehicles are the same at \( t = \frac{3}{4} \) hours (30 km/hr) and \( t = 1.5 \) hours (60 km/hr).
2. The second car does not overtake the first car during the journey.