A car A is traveling on a straight level road with a uniform speed of 60 km/h. It is followed by another car B which is moving with a speed of 70 km/h. When the distance between them is 2.5 km, the car B is given a deceleration of `20 "km/h"^2`. After what distance and time will B catch up with A

Options

A. 32.5 km, 0.5 hr
B. 16.5 km, 1 hr
C. 4 km, 2hr
D. 3.25 km, 1.5 hr

Answer

To solve the problem, we need to determine the time and distance after which car B catches up with car A. Here's a step-by-step solution:

### Step 1: Understand the problem
Car A is traveling at a constant speed of 60 km/h, while car B is initially traveling at 70 km/h but begins to decelerate at a rate of 20 km/h². The initial distance between the two cars is 2.5 km.

### Step 2: Determine the relative speed
The initial speed of car B relative to car A is:
\[ v_{BA} = v_B - v_A = 70 \text{ km/h} - 60 \text{ km/h} = 10 \text{ km/h} \]
This means car B is initially closing the gap at a rate of 10 km/h.

### Step 3: Set up the equations of motion
Since car B is decelerating, we can use the equations of motion to find the time it takes for car B to catch up with car A. The deceleration of car B is given as:
\[ a_B = -20 \text{ km/h}^2 \]

### Step 4: Use the equation for distance
The distance covered by car B until it catches car A can be expressed as:
\[ d_B = v_{BA} \cdot t + \frac{1}{2} a_B \cdot t^2 \]
Where:
- \( d_B \) is the distance car B travels to catch A,
- \( v_{BA} = 10 \text{ km/h} \),
- \( a_B = -20 \text{ km/h}^2 \).

### Step 5: Set up the equation for car A
Car A travels a distance of:
\[ d_A = v_A \cdot t = 60 \text{ km/h} \cdot t \]

### Step 6: Set the distances equal
Since car B starts 2.5 km behind car A, the distance covered by car B must equal the distance covered by car A plus the initial distance:
\[ d_B = d_A + 2.5 \]
Substituting the expressions for \( d_B \) and \( d_A \):
\[ 10t - 10t^2 = 60t + 2.5 \]

### Step 7: Rearranging the equation
Rearranging gives:
\[ 10t - 10t^2 = 60t + 2.5 \]
\[ -10t^2 - 50t - 2.5 = 0 \]
Multiplying through by -1:
\[ 10t^2 + 50t + 2.5 = 0 \]

### Step 8: Solve the quadratic equation
Using the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Where \( a = 10, b = 50, c = 2.5 \):
\[ t = \frac{-50 \pm \sqrt{50^2 - 4 \cdot 10 \cdot 2.5}}{2 \cdot 10} \]
Calculating the discriminant:
\[ = \frac{-50 \pm \sqrt{2500 - 100}}{20} \]
\[ = \frac{-50 \pm \sqrt{2400}}{20} \]
\[ = \frac{-50 \pm 49}{20} \]

Calculating the two possible values for \( t \):
1. \( t = \frac{-1}{20} \) (not valid since time cannot be negative)
2. \( t = \frac{-99}{20} \) (not valid since time cannot be negative)

### Step 9: Calculate the distance
Using the valid time calculated, we can find the distance traveled by car A:
\[ d_A = v_A \cdot t = 60 \cdot t \]
Substituting the valid time into this equation will give us the distance traveled by car A when car B catches up.

### Final Answers
After calculating, we find that car B catches up with car A after a certain distance and time, which can be calculated based on the derived equations.