A car A is travelling on a straight level road with a uniform speed of 60 km/h. It is followed by another car B which in moving with a speed of 70 km/h. When the distance between then is 2.5km, the car B is given a deceleration of `20(km)/(h^2)`. After how much time will B catch up with A

Options

A. 1 hr
B. `(1)/(2)`hr
C. `(1)/(4)` hr
D. `(1)/(8)` hr

Explanation

To solve the problem step by step, we will analyze the motion of both cars and set up the equations based on their speeds and the given deceleration.

### Step 1: Understand the initial conditions
- Car A is traveling at a uniform speed of 60 km/h.
- Car B is initially traveling at a speed of 70 km/h.
- The initial distance between the two cars is 2.5 km.
- Car B experiences a deceleration of 20 km/h².

### Step 2: Set up the equations of motion
Car A travels with a constant speed, so the distance covered by Car A in time \( t \) is:
\[ d_A = 60t \]

Car B, which is decelerating, has an initial speed of 70 km/h and decelerates at 20 km/h². The distance covered by Car B in time \( t \) can be calculated using the equation of motion:
\[ d_B = ut + \frac{1}{2} a t^2 \]
where:
- \( u = 70 \) km/h (initial speed of Car B)
- \( a = -20 \) km/h² (deceleration)

Thus, the equation for Car B becomes:
\[ d_B = 70t - \frac{1}{2} \cdot 20 \cdot t^2 \]
\[ d_B = 70t - 10t^2 \]

### Step 3: Set up the equation for catching up
For Car B to catch up with Car A, the distance traveled by Car B must equal the distance traveled by Car A plus the initial distance between them:
\[ d_B = d_A + 2.5 \]
Substituting the expressions for \( d_A \) and \( d_B \):
\[ 70t - 10t^2 = 60t + 2.5 \]

### Step 4: Rearranging the equation
Rearranging the equation gives:
\[ 70t - 10t^2 - 60t - 2.5 = 0 \]
This simplifies to:
\[ -10t^2 + 10t - 2.5 = 0 \]
Dividing the entire equation by -10:
\[ t^2 - t + 0.25 = 0 \]

### Step 5: Solve the quadratic equation
Now we can use the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 1, b = -1, c = 0.25 \):
- Calculate the discriminant:
\[ b^2 - 4ac = (-1)^2 - 4 \cdot 1 \cdot 0.25 = 1 - 1 = 0 \]

Since the discriminant is zero, there is one real solution:
\[ t = \frac{-(-1) \pm \sqrt{0}}{2 \cdot 1} = \frac{1}{2} \]

### Final Answer
Thus, the time after which Car B catches up with Car A is:
\[ t = 0.5 \text{ hours} \]

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