Two cars A and B are moving uniformly on two straight roads at right angles to one another at 40 and 20 km/h respectively. A passes the intersection of the road when B has still to move 50 km to reach it. The shortest distance between the two cars and the time when they are closet

Options

A. `20 sqrt"" 5 km., 30 m `
B. 20 km., 10
C. 25 km., 20
D. none of these

Answer

To solve the problem, we need to find the shortest distance between two cars A and B moving at right angles to each other and the time when they are closest. Let's break down the solution step by step.

### Step 1: Understand the scenario
Car A is moving at a speed of 40 km/h and car B is moving at a speed of 20 km/h. When car A passes the intersection, car B is still 50 km away from it.

### Step 2: Set up the coordinate system
Let's set the intersection point as the origin (0, 0) of our coordinate system.
- Car A moves along the x-axis.
- Car B moves along the y-axis.

At time \( t = 0 \):
- Position of car A: \( (0, 0) \)
- Position of car B: \( (0, -50) \) (since it is 50 km away from the intersection)

### Step 3: Write the equations of motion
The position of car A at time \( t \) (in hours) is given by:
\[ x_A = 40t \]
\[ y_A = 0 \]

The position of car B at time \( t \) is given by:
\[ x_B = 0 \]
\[ y_B = -50 + 20t \]

### Step 4: Calculate the distance between the two cars
The distance \( d \) between the two cars at time \( t \) can be calculated using the distance formula:
\[ d = \sqrt{(x_A - x_B)^2 + (y_A - y_B)^2} \]
Substituting the positions:
\[ d = \sqrt{(40t - 0)^2 + (0 - (-50 + 20t))^2} \]
\[ d = \sqrt{(40t)^2 + (50 - 20t)^2} \]

### Step 5: Simplify the distance equation
Expanding the equation:
\[ d = \sqrt{1600t^2 + (50 - 20t)^2} \]
\[ d = \sqrt{1600t^2 + 2500 - 2000t + 400t^2} \]
\[ d = \sqrt{2000t^2 - 2000t + 2500} \]

### Step 6: Find the time when the distance is minimized
To find the minimum distance, we can take the derivative of \( d^2 \) (to avoid dealing with the square root) and set it to zero:
Let \( D = d^2 = 2000t^2 - 2000t + 2500 \).

Taking the derivative:
\[ \frac{dD}{dt} = 4000t - 2000 \]
Setting the derivative to zero:
\[ 4000t - 2000 = 0 \]
\[ 4000t = 2000 \]
\[ t = \frac{1}{2} \text{ hours} = 30 \text{ minutes} \]

### Step 7: Calculate the minimum distance
Substituting \( t = \frac{1}{2} \) back into the distance equation:
\[ d = \sqrt{2000\left(\frac{1}{2}\right)^2 - 2000\left(\frac{1}{2}\right) + 2500} \]
\[ d = \sqrt{2000 \cdot \frac{1}{4} - 1000 + 2500} \]
\[ d = \sqrt{500 - 1000 + 2500} \]
\[ d = \sqrt{2000} \]
\[ d = 20\sqrt{5} \text{ km} \]

### Final Answers
- The shortest distance between the two cars is \( 20\sqrt{5} \) km.
- The time when they are closest is 30 minutes.