MHB Eb5 Calculate the average speed in km/h

karush
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$\text{A person jogs eight complete laps around a $400 \, m$ track in a total time of $14.5 \, min$}. \\$
$\textit{a. Calculate the average speed in $km/h$}\\$
\begin{align*}\displaystyle \frac{(8)400 \, m}{14.5 \, min} \left(\frac{km}{1000m} \right) \left(\frac{60 \, min}{h} \right)&=\frac{192 \ km}{14.5 \, h}\\ &\approx\color{red}{13.2 \, km/h}\end{align*}

left out some steps but hope answer is correct
next question is average velocity
 
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karush said:
$\text{A person jogs eight complete laps around a $400 \, m$ track in a total time of $14.5 \, min$}. \\$
$\textit{a. Calculate the average speed in $km/h$}\\$
\begin{align*}\displaystyle \frac{(8)400 \, m}{14.5 \, min} \left(\frac{km}{1000m} \right) \left(\frac{60 \, min}{h} \right)&=\frac{192 \ km}{14.5 \, h}\\ &\approx\color{red}{13.2 \, km/h}\end{align*}

left out some steps but hope answer is correct
next question is average velocity
Yes, that is correct. The average velocity is 0! Do you see why? What is the distinction between "speed" and "velocity"?
 
well i thot $\textsf{average velocity, straight-line motion was}$\begin{align*}\displaystyle v_{av}&=\frac{x_2-x_1}{t_2-t_1}=\frac{\Delta x}{\Delta t}\end{align*}so we plug in what?
 

But this is NOT "straight line motion"! Again, what is the difference between "speed" and "velocity"?
 
HallsofIvy said:

But this is NOT "straight line motion"! Again, what is the difference between "speed" and "velocity"?

but isn't the ball being thrown strsight up and then fall straight down
 
karush said:
but isn't the ball being thrown strsight up and then fall straight down
Uh, no this is about a person running about an oval track! You seem to have two threads confused.
 
Ok
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