Question

solve the following (show your work): 1. maddix drives a car that is uniformly accelerated at the rate of 2.5m/s² for 12s. if the original speed of the car is 8.0m/s, what is its final velocity? 2. jesse is in a car whose velocity increases from rest to 14 m/s, in 3.5 sec. what is its acceleration? what is its acceleration if it next slows down to 7 m/s, in 2 sec? 3. emily is in a car that is moving at 16.67 m/s when it begins to decelerate at 1.5m/s². how long does it take for it to slow down to 8.7 m/s?

Explanation:

Step1: Identify the formula for final - velocity in uniformly - accelerated motion

The formula is $v = v_0+at$, where $v$ is the final velocity, $v_0$ is the initial velocity, $a$ is the acceleration, and $t$ is the time.
Given $v_0 = 8.0m/s$, $a = 2.5m/s^2$, and $t = 12s$.
$v=8.0 + 2.5\times12$

Step2: Calculate the final velocity

$v=8.0+30$
$v = 38m/s$

Step3: For the second - part, find the acceleration when velocity changes from rest ($v_0 = 0m/s$) to $v = 14m/s$ in $t = 3.5s$

The acceleration formula is $a=\frac{v - v_0}{t}$.
$a=\frac{14 - 0}{3.5}$
$a = 4m/s^2$

Step4: Find the acceleration when the car slows down from $v_0 = 14m/s$ to $v = 7m/s$ in $t = 2s$

Using $a=\frac{v - v_0}{t}$, we have $a=\frac{7 - 14}{2}$
$a=-3.5m/s^2$

Step5: For the third - part, find the time when the car decelerates

The formula $v = v_0+at$ can be rewritten for time as $t=\frac{v - v_0}{a}$. Here, $v_0 = 16.67m/s$, $v = 8.7m/s$, and $a=-1.5m/s^2$ (negative because it's deceleration).
$t=\frac{8.7 - 16.67}{-1.5}$
$t=\frac{-7.97}{-1.5}\approx5.31s$

Answer:

1. The final velocity of Maddix's car is $38m/s$.
2. The acceleration when the car speeds up is $4m/s^2$, and the acceleration when it slows down is $-3.5m/s^2$.
3. It takes approximately $5.31s$ for Emily's car to slow down.