Two straight roads OA and OB intersect at an angle `60^@. A ` car approaches O from A, where `OA = 700 m` at a uniform speed of 20 m/s, Simultaneously, a runner starts running from O towards B at a uniform speed of 5 m/s. The time after start when the car and the runner are closest is :

Options

A. 10 sec
B. 15 sec
C. 20 sec
D. 30 sec

Answer

To solve the problem step by step, we will analyze the positions of the car and the runner over time, and then find the time when they are closest to each other.

### Step 1: Understand the Setup
- The car is approaching point O from point A, which is 700 m away, at a speed of 20 m/s.
- The runner is starting from point O and running towards point B at a speed of 5 m/s.
- The angle between the roads OA and OB is 60 degrees.

### Step 2: Define the Positions
- Let \( t \) be the time in seconds after both the car and the runner start moving.
- The position of the car after time \( t \) is:
\[ \text{Position of Car} = 700 - 20t \]
- The position of the runner after time \( t \) is:
\[ \text{Position of Runner} = 5t \]

### Step 3: Determine the Coordinates
- The coordinates of the car (C) after time \( t \) can be represented as:
\[ C = (700 - 20t, 0) \]
- The coordinates of the runner (R) can be represented as:
\[ R = (5t \cos 60^\circ, 5t \sin 60^\circ) = (5t \cdot \frac{1}{2}, 5t \cdot \frac{\sqrt{3}}{2}) = \left(\frac{5t}{2}, \frac{5\sqrt{3}t}{2}\right) \]

### Step 4: Calculate the Distance
- The distance \( D \) between the car and the runner can be calculated using the distance formula:
\[ D^2 = \left(\frac{5t}{2} - (700 - 20t)\right)^2 + \left(\frac{5\sqrt{3}t}{2} - 0\right)^2 \]
- Simplifying the first term:
\[ D^2 = \left(\frac{5t}{2} + 20t - 700\right)^2 + \left(\frac{5\sqrt{3}t}{2}\right)^2 \]
\[ = \left(\frac{5t + 40t - 1400}{2}\right)^2 + \left(\frac{5\sqrt{3}t}{2}\right)^2 \]
\[ = \left(\frac{45t - 1400}{2}\right)^2 + \left(\frac{5\sqrt{3}t}{2}\right)^2 \]

### Step 5: Differentiate to Find Minimum Distance
- To find the time when the distance is minimized, we differentiate \( D^2 \) with respect to \( t \) and set it to zero:
\[ D^2 = \left(\frac{45t - 1400}{2}\right)^2 + \left(\frac{5\sqrt{3}t}{2}\right)^2 \]
- Let \( u = \frac{45t - 1400}{2} \) and \( v = \frac{5\sqrt{3}t}{2} \):
\[ D^2 = u^2 + v^2 \]
- Differentiate:
\[ \frac{d(D^2)}{dt} = 2u \cdot \frac{du}{dt} + 2v \cdot \frac{dv}{dt} = 0 \]
- Calculate \( \frac{du}{dt} \) and \( \frac{dv}{dt} \):
\[ \frac{du}{dt} = \frac{45}{2}, \quad \frac{dv}{dt} = \frac{5\sqrt{3}}{2} \]
- Setting the derivative to zero gives:
\[ 2\left(\frac{45t - 1400}{2}\right) \cdot \frac{45}{2} + 2\left(\frac{5\sqrt{3}t}{2}\right) \cdot \frac{5\sqrt{3}}{2} = 0 \]

### Step 6: Solve for \( t \)
- Solving the equation leads us to:
\[ 45(45t - 1400) + 75t = 0 \]
- Rearranging gives:
\[ 2025t - 63000 + 75t = 0 \]
\[ 2100t = 63000 \implies t = \frac{63000}{2100} = 30 \text{ seconds} \]

### Conclusion
The time after which the car and the runner are closest is **30 seconds**.