MHB Eb3 Velocity and Time: Calculate $V_{av}, v_{av}$, and Time to Fly $3.5 \ km$

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To calculate the average velocity of a car traveling 235 km in 2.75 hours, the result is approximately 85.45 km/h. For a particle moving from 4.8 cm at -2.0 s to 8.5 cm at 4.5 s, the average velocity is about 0.57 cm/s, despite the negative time value. A bird flying at 25 km/h takes approximately 8.4 minutes to cover 3.5 km. The discussion emphasizes the need for clear formatting in problem-solving, akin to textbook solutions. Overall, the calculations are confirmed as correct, with suggestions to include units for clarity.
karush
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1 What must your car’s $V_{av}$ be in order to travel $\textbf{235 km}$ in $\textbf{2.75 h}$
\begin{align*}\displaystyle
V_{av}&=235km/2.75h \, km/h \\
&=\color{red}{85.45 \, km/h}
\end{align*}

2 A particle at $t_1 =–2.0 s$ is at $x_1 = 4.8 cm$ and at $t_2 = 4.5 s$ is at $x_2 = 8.5 cm$.}\\
i. What is its average velocity over this time interval?\\
ii. Can this be calculated. Why or why not?
\begin{align*}\displaystyle
v_{av}&=\frac{x_2-x_1} {t_2-t_1}\\
&=\frac{8.5-(4.8)}{4.5-(-2.0)}\\
&\approx \color{red}{.57}
\end{align*}

3 A bird can fly $25 km/h$.
How long does it take to fly $3.5 km?$\begin{align*}\displaystyle
D&=R \cdot T\\
\therefore \frac{D}{R}&=T \\
\frac{25}{60}=\frac{5}{12}&=\frac{3.5}{min}\\
5 \ min&=3.5(12) \\
T&=.7(12)=\color{red}{8.4 \, min}
\end{align*}

Ok I know these are relatively easy problems but they will get harder fast,
there is no book anwswer this is from a class that is already over.
I am trying to format these problems so that they look like a math textbook solution
So suggestions are very welcome.
#2 has a negative time in it but I assumed it could be calculated anyway$$\tiny\textit{Embry-Riddle Aeronautical University Dept of Physical Sciences, HW $\# 1-3$ PS 103 / Technical Physics I}$$
 
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1. Correct
2. Correct, might want to include units.
3. Correct.
 
Rido12 said:
1. Correct
2. Correct, might want to include units.
3. Correct.

$\begin{align*}\displaystyle
v_{av}&=\frac{x_2-x_1} {t_2-t_1}\\
&=\frac{8.5-(4.8)}{4.5-(-2.0)}\\
&\approx \color{red}{.57 \, cm/s}
\end{align*}$
 
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