# A car traveling at speed v takes distance d to stop after the brakes are applied. What is the stopping distance if the car is initially traveling at speed 7.0v? Assume that the acceleration due to the braking is the same in both cases.

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.

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.