This problem aims to find the **distance** the car covers with **negative acceleration** when its brakes are applied. This problem requires the understanding of basic applied physics including **velocity**, **acceleration**, and the **three equations of motions**.

We can define **deceleration** as the opposite or negative of acceleration. This deceleration can be calculated by dividing the difference between **final velocity** $v_f$ and the **initial velocity** $v_i$ by the amount of time $t$ it takes to lower its velocity. The formula for deceleration is the same as that of acceleration but with a **negative** **sign**, which is helpful in determining the value of deceleration.

## Expert Answer

In applied physics, we use the **equations of motion** to determine the behavior of a physical system when there is a motion of an object as a function of **time**. More precisely, the equations of motion define the conduct of a physical approach as a group of **mathematical functions** in terms of dynamic variables.

Using the **third equation** of motion:

\[ v^2 = u^2 + 2ad \hspace {3ex} …… equation(1) \]

where:

$a$ = **acceleration**

**$u$ = initial velocity**

**$v$ = final velocity**

**$d$ = distance travelled**

When brakes are applied the car starts to** slow down** until its velocity reaches $0$, so we can put the final velocity $v$ equals to $0$,

\[ 0 = u^2 + 2ad\]

\[ u^2 = -2ad\]

From here, we can rearrange the formula to determine the value of **acceleration** $a$:

\[ a = \left( – \dfrac{u^2} {2d} \right) \hspace {3ex} …… equation(2) \]

Now putting the expression of $a$ from $equation (2)$ into the $equation(1)$ above, where the **final velocity** $v$ equals to $0$ and $7v$ is the initial velocity $u$.

\[ 0 = (7.0v)^2 + 2 \left( – \dfrac{v^2}{2d}\right) d’ \]

$d’$ is the **stopping** distance that we are looking for:

\[ 2 \left( \dfrac{v^2} {2d}\right) d’ = (7.0v)^2 \]

\[ \left( \dfrac{v^2} {d} \right) d’ = 49.0 v^2 \]

\[ v^2 d’ = 49.0 v^2d \]

\[ d’ = 49.0 d \]

## Numerical Result

Hence, the car’s **stopping distance** which is initially traveling at a speed of $7.0v$ is $49d$.

## Example

A car traveling with a speed of $72km/h$ applies its brakes. What is the stopping **distance** if it experiences constant **retardation** of $40m/s^2$?

The **initial velocity** of the car is $72 km/h$, converting it into $m/s$ gives us $20 m/s$.

As the **retardation** is in the **opposite direction** to the initial velocity of the car, the **acceleration** $a$ becomes $-40 m/s^2$.

The **final velocity** of the car is given as $0 m/s$.

Using the** third equation of motion** to find the stopping distance in which the car stops when the brakes are applied:

\[v^2 – u^2 = 2as\]

Substituting the values to solve for $s$:

\[ 0^2 – 20^2 = 2 (-40) s \]

\[ -400 = -90s \]

\[ s = 5m \]

The **stopping distance** at which the car stops when the breaks are applied given the initial velocity of the car was $72km/h$ comes out to be $s = 5$ meters.