A car is moving towards south with a speed of `20 m s^(-1)`. A motorcycst is moving towards east with a speed of `15 m s^(-1)`. At a crttain instant, the motorcyclistis due south of the car and is at a distance of `50 m`from the car.
The shortest distance between the motorcyclist and the car is.

Options

A. a. `20 m`
B. b.`10 m`
C. c.`40 m`
D. d. `30 m`

Explanation

To solve the problem of finding the shortest distance between a car moving south and a motorcyclist moving east, we can follow these steps:

### Step 1: Understand the Initial Setup
- The car is moving south at a speed of \( V_c = 20 \, \text{m/s} \).
- The motorcyclist is moving east at a speed of \( V_m = 15 \, \text{m/s} \).
- At the initial moment, the distance between the car and the motorcyclist is \( d = 50 \, \text{m} \).

### Step 2: Set Up the Position Functions
- Let the position of the car at time \( t \) be \( y_c(t) = -20t \) (moving south).
- Let the position of the motorcyclist at time \( t \) be \( x_m(t) = 15t \) (moving east).
- The initial position of the car can be considered at the origin (0, 0) and the motorcyclist at (0, -50).

### Step 3: Find the Distance Function
- The distance \( S \) between the car and the motorcyclist at time \( t \) can be expressed using the distance formula:
\[ S(t) = \sqrt{(x_m(t) - 0)^2 + (y_c(t) + 50)^2} \]
Substituting the position functions:
\[ S(t) = \sqrt{(15t)^2 + (-20t + 50)^2} \]
Simplifying this:
\[ S(t) = \sqrt{(15t)^2 + (50 - 20t)^2} \]
\[ S(t) = \sqrt{225t^2 + (50 - 20t)^2} \]
\[ S(t) = \sqrt{225t^2 + (2500 - 2000t + 400t^2)} \]
\[ S(t) = \sqrt{625t^2 - 2000t + 2500} \]

### Step 4: Minimize the Distance
- To find the shortest distance, we need to minimize \( S(t) \). We can do this by finding the derivative \( \frac{dS}{dt} \) and setting it to zero:
\[ \frac{dS}{dt} = \frac{1}{2\sqrt{625t^2 - 2000t + 2500}} \cdot (1250t - 2000) = 0 \]
- Setting the numerator to zero:
\[ 1250t - 2000 = 0 \implies t = \frac{2000}{1250} = 1.6 \, \text{s} \]

### Step 5: Calculate the Shortest Distance
- Now substitute \( t = 1.6 \) seconds back into the distance formula:
\[ S(1.6) = \sqrt{625(1.6)^2 - 2000(1.6) + 2500} \]
\[ S(1.6) = \sqrt{625 \cdot 2.56 - 3200 + 2500} \]
\[ S(1.6) = \sqrt{1600 - 3200 + 2500} \]
\[ S(1.6) = \sqrt{900} = 30 \, \text{m} \]

### Final Answer
The shortest distance between the motorcyclist and the car is \( 30 \, \text{m} \).

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