Question

1. a baseball is thrown upward and slows from 20 m/s to 5 m/s in 1.5 seconds. what is its acceleration? solve for time (t) 7. a car slows down from 40 m/s to 20 m/s with an acceleration of -5.0 m/s². how long does this take? 8. a skateboarder decelerates at -2.0 m/s², going from 12 m/s to 0 m/s. how much time passes during the acceleration? solve for change in velocity (δv) 9. a truck accelerates at -1.2 m/s² for 10 seconds. what is the change in velocity? 10. a sprinter decelerates at -3.0 m/s² for 3.0 seconds. what is the change in velocity?

Explanation:

Step1: Recall acceleration formula

The formula for acceleration is $a=\frac{\Delta v}{\Delta t}=\frac{v_f - v_i}{t}$, where $v_f$ is the final - velocity, $v_i$ is the initial - velocity, and $t$ is the time.

Step2: Solve problem 6

Given $v_i = 20\ m/s$, $v_f = 5\ m/s$, and $t = 1.5\ s$.
$a=\frac{v_f - v_i}{t}=\frac{5 - 20}{1.5}=\frac{- 15}{1.5}=-10\ m/s^2$

Step3: Solve problem 7

We know $a=\frac{v_f - v_i}{t}$, and we want to solve for $t$. Rearranging the formula gives $t=\frac{v_f - v_i}{a}$.
Given $v_i = 40\ m/s$, $v_f = 20\ m/s$, and $a=-5.0\ m/s^2$.
$t=\frac{20 - 40}{-5.0}=\frac{-20}{-5.0}=4\ s$

Step4: Solve problem 8

Using $t=\frac{v_f - v_i}{a}$, with $v_i = 12\ m/s$, $v_f = 0\ m/s$, and $a=-2.0\ m/s^2$.
$t=\frac{0 - 12}{-2.0}=\frac{-12}{-2.0}=6\ s$

Step5: Recall the formula for change in velocity

The formula for change in velocity is $\Delta v=a\times t$.

Step6: Solve problem 9

Given $a=-1.2\ m/s^2$ and $t = 10\ s$.
$\Delta v=a\times t=-1.2\times10=-12\ m/s$

Step7: Solve problem 10

Given $a=-3.0\ m/s^2$ and $t = 3.0\ s$.
$\Delta v=a\times t=-3.0\times3.0=-9.0\ m/s$

Answer:

1. $-10\ m/s^2$
2. $4\ s$
3. $6\ s$
4. $-12\ m/s$
5. $-9.0\ m/s$