Question

1. a car traveling at a speed of 30.0 m/s encounters an emergency and comes to a complete stop. how much time will it take for the car to stop if it decelerates at -4.0 m/s²? 5. if a car can go from 9 to 26.82 m/s in 8 seconds, what would be its final speed after 5.0 seconds if its starting speed were 22.352 m/s?

Explanation:

Step1: Identify the kinematic - equation

We use the kinematic equation $v = v_0+at$, where $v$ is the final velocity, $v_0$ is the initial velocity, $a$ is the acceleration, and $t$ is the time.

Step2: Substitute values for problem 4

Given $v_0 = 30.0\ m/s$, $a=- 4.0\ m/s^2$, and the final velocity $v = 0\ m/s$. We want to find the time $t$. Rearranging the equation $v = v_0+at$ for $t$ gives $t=\frac{v - v_0}{a}$.
Substitute the values: $t=\frac{0 - 30.0}{-4.0}=\frac{-30.0}{-4.0}=7.5\ s$.

Step3: Substitute values for problem 5

Given $v_0 = 22.352\ m/s$, $a=\frac{26.82 - 0}{8}=3.3525\ m/s^2$ (first - find the acceleration from the initial data where $v_0 = 0$, $v = 26.82\ m/s$ and $t = 8\ s$ using $a=\frac{v - v_0}{t}$), and $t = 5.0\ s$.
Using the equation $v = v_0+at$, we substitute $v_0 = 22.352\ m/s$, $a = 3.3525\ m/s^2$, and $t = 5.0\ s$.
$v=22.352+3.3525\times5=22.352 + 16.7625=39.1145\ m/s$.

Answer:

1. The time it takes for the car to stop is $7.5\ s$.
2. The final speed of the car is $39.1145\ m/s$.