The driver of a vehicle moving with a speed of `72 km h^(-1)` sees a child standing in the middle of the road. She applies the brakes and brings the vehicle to rest just in time to save the child. If the time taken to stop the vehicle is 10 s and the mass of the driver and the vehicle is 500 kg, then the average retarding force on the vehicle is A.

Options

A. `1 xx 10^(8) g cm s^(-2)`
B. `1 xx 10^(4) g m s^(2)`
C. `1 xx 10^(6) kg cm s^(-2)`
D. ` 1 xx 10^(2) kg m s^(-2)`

Answer

To solve the problem, we need to find the average retarding force acting on the vehicle. We can follow these steps:

### Step 1: Convert the speed from km/h to m/s
The initial speed of the vehicle is given as 72 km/h. To convert this to meters per second (m/s), we use the conversion factor:
\[ 1 \text{ km/h} = \frac{1}{3.6} \text{ m/s} \]
Thus,
\[ 72 \text{ km/h} = \frac{72}{3.6} \text{ m/s} = 20 \text{ m/s} \]

### Step 2: Calculate the acceleration (deceleration)
The vehicle comes to rest, which means its final velocity (v) is 0 m/s. The initial velocity (u) is 20 m/s, and the time (t) taken to stop is 10 seconds. We can use the formula for acceleration:
\[ a = \frac{v - u}{t} \]
Substituting the values:
\[ a = \frac{0 - 20}{10} = \frac{-20}{10} = -2 \text{ m/s}^2 \]
(Note: The negative sign indicates deceleration.)

### Step 3: Calculate the average retarding force
We can use Newton's second law to find the force. The formula is:
\[ F = m \cdot a \]
where:
- \( F \) is the force,
- \( m \) is the mass of the vehicle and driver (500 kg),
- \( a \) is the acceleration (-2 m/s²).

Substituting the values:
\[ F = 500 \cdot (-2) = -1000 \text{ N} \]
The negative sign indicates that the force is acting in the opposite direction to the motion (retarding force).

### Final Answer
The average retarding force on the vehicle is **1000 N**.

---