Two highways are perpendicular to each other `:` imagine them to be along the x-axis and the y-axis, respectively. At the instant t=0, a police car P is at a distance d= 400m from the intersection and moving at speed of 80km`//`h towards it along the x-axis. Motorist M is at a distance of 600m from the intersection and moving towards it at a speed of 60km`//`h along the y-axis . The minimum distance between the cars is

Answer

To solve the problem of finding the minimum distance between the police car P and the motorist M, we can follow these steps:

### Step 1: Define the initial positions and velocities
- The police car P is initially at a distance of \( d_P = 400 \, \text{m} = 0.4 \, \text{km} \) from the intersection, moving towards it along the x-axis at a speed of \( v_P = 80 \, \text{km/h} \).
- The motorist M is initially at a distance of \( d_M = 600 \, \text{m} = 0.6 \, \text{km} \) from the intersection, moving towards it along the y-axis at a speed of \( v_M = 60 \, \text{km/h} \).

### Step 2: Write the equations for the positions of the cars over time
At time \( t \) (in hours):
- The position of the police car P along the x-axis is given by:
\[ x_P(t) = 0.4 - 80t \]
- The position of the motorist M along the y-axis is given by:
\[ y_M(t) = 0.6 - 60t \]

### Step 3: Write the expression for the distance between the two cars
The distance \( D(t) \) between the two cars at time \( t \) can be expressed using the Pythagorean theorem:
\[ D(t) = \sqrt{(x_P(t))^2 + (y_M(t))^2} \]
Substituting the expressions for \( x_P(t) \) and \( y_M(t) \):
\[ D(t) = \sqrt{(0.4 - 80t)^2 + (0.6 - 60t)^2} \]

### Step 4: Simplify the expression for distance
Squaring the distance to simplify calculations:
\[ D^2(t) = (0.4 - 80t)^2 + (0.6 - 60t)^2 \]
Expanding both squares:
\[ D^2(t) = (0.16 - 64t + 6400t^2) + (0.36 - 72t + 3600t^2) \]
Combining like terms:
\[ D^2(t) = 0.52 - 136t + 10000t^2 \]

### Step 5: Minimize the distance
To find the minimum distance, we take the derivative of \( D^2(t) \) with respect to \( t \) and set it to zero:
\[ \frac{d(D^2)}{dt} = -136 + 20000t = 0 \]
Solving for \( t \):
\[ 20000t = 136 \implies t = \frac{136}{20000} = 0.0068 \, \text{hours} \]

### Step 6: Calculate the minimum distance
Now we substitute \( t \) back into the expression for \( D(t) \):
\[ D^2(0.0068) = 0.52 - 136(0.0068) + 10000(0.0068)^2 \]
Calculating each term:
1. \( 0.52 - 136 \times 0.0068 = 0.52 - 0.9248 = -0.4048 \)
2. \( 10000 \times (0.0068)^2 = 10000 \times 0.00004624 = 0.4624 \)

Combining these:
\[ D^2(0.0068) = -0.4048 + 0.4624 = 0.0576 \]
Thus, the minimum distance \( D \) is:
\[ D = \sqrt{0.0576} \approx 0.24 \, \text{km} = 240 \, \text{m} \]

### Final Answer
The minimum distance between the police car and the motorist is **240 meters**.

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