a driver of a car traveling at 15.0 m/s applies the brakes, causing a uniform acceleration of - 2.0 m/s²…

Explanation:

Step1: Identify the known - values

Initial velocity $u = 15.0\ m/s$, final velocity $v = 10.0\ m/s$, acceleration $a=- 2.0\ m/s^{2}$.

Step2: Use the kinematic equation $v = u+at$ to find the time $t$

We can re - arrange the equation $v = u + at$ to solve for $t$. So, $t=\frac{v - u}{a}$. Substituting the values: $t=\frac{10.0 - 15.0}{-2.0}=\frac{-5.0}{-2.0}=2.5\ s$.

Step3: Use the kinematic equation $v^{2}-u^{2}=2ax$ to find the distance $x$

We re - arrange the equation $v^{2}-u^{2}=2ax$ to solve for $x$. So, $x=\frac{v^{2}-u^{2}}{2a}$. Substituting $u = 15.0\ m/s$, $v = 10.0\ m/s$ and $a=-2.0\ m/s^{2}$: [ \begin{align*} x&=\frac{10.0^{2}-15.0^{2}}{2\times(-2.0)}\ &=\frac{100 - 225}{-4.0}\ &=\frac{-125}{-4.0}\ & = 31.25\ m \end{align*} ]

Answer:

The time it takes for the car to decelerate is $2.5\ s$ and the distance the car has moved during the braking is $31.25\ m$.