Answer :

Answer:

28.57

Explanation:

To find the distance the car travels before it stops, we can use the equations of motion for uniformly decelerated motion.

Given:

- Initial velocity, \( u = 20 \) m/s (since the car is slowing down, we take it as positive)

- Acceleration (deceleration), \( a = -7 \) m/s\(^2\) (negative because it's deceleration)

- Final velocity, \( v = 0 \) m/s (since the car stops)

We need to find the distance \( s \).

The equation relating initial velocity, final velocity, acceleration, and distance is:

\[ v^2 = u^2 + 2as \]

Substitute the given values:

\[ 0 = (20)^2 + 2 \cdot (-7) \cdot s \]

Simplify and solve for \( s \):

\[ 0 = 400 - 14s \]

\[ 14s = 400 \]

\[ s = \frac{400}{14} \]

\[ s = 28.57 \text{ meters} \]

So, the car travels approximately \( 28.57 \) meters before it stops.