MHB Driving San Diego to LA: 1h 10min Longer on a Fri Afternoon

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Driving from San Diego to Los Angeles typically takes 2 hours and 20 minutes at an average speed of 65 mph. However, on a Friday afternoon, heavy traffic reduces the speed to 43 mph, increasing the travel time to approximately 3 hours and 30 minutes. The difference in travel time is about 1 hour and 10 minutes longer due to the slower speed. The calculations confirm that the distance remains constant, and the use of units is correctly applied throughout the process. Overall, traffic significantly impacts travel duration on this route.
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2-3 You normally drive on the freeway between San Diego and Los Angel es at an average speed of
$105 \, km/h \, (65 \, mi/h)$ and the trip takes $2h$ and $20 \, min$
On a Friday afternoon, however, heavy traffic slows you down and you drive the same distance at an average speed of only $70 km/h \, (43 mi/h)$
How much longer does it take?
\begin{align*}
d_1&=r_1t_1=
\dfrac{65\,mi}{\cancel{ h }}
\cdot 2.33\, \cancel{h}=151.45\,mi\\
t_2&=\dfrac{d_1}{r_2}=
151.45\,\cancel{mi}\cdot\dfrac{h}{43\,\cancel{mi}}=3.52\,h\\
t_\Delta&=t_2-t_1=
3.52h-2.33h
\approx 1.19\, h \approx 1\, h\, 10\, min
\end{align*}ok I think this is correct but was not sure about the units how to use them in process
since there is 2 sets of time and rate.
also this cancel control kinda sucks should be same angle or straight and red maybe.
 
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Yes, if you drive at 65 mi/hr for 2 and 1/3 hours, then you have driven (65)(7/3)= 151 and 2/3 miles. That is the distance from San Francisco to Los Angeles. Driving that same distance at 42 mi/hr, will take a time of 151 and 2/3 divided by 42= 3 and 25/42 hours.

Notice that, to two decimal places, 25/42= 0.59 not 0.52. I prefer not to round until the end, if[/b that is necessary. Here, the data was given as a fraction so I would think it better to leave the answer as a fraction: 3 and 25/42 minus 2 and 1/3= 3 and 25/42- 2 and 14/42= 1 and 11/42 hours.

Yes, your use of the "units" is correct.
 
$r_1t_1 = r_2t_2 \implies t_2 = \dfrac{r_1t_1}{r_2} = \dfrac{(105 \, km/hr)(7/3 \, hr)}{70 \, km/hr} = \dfrac{3}{2} \cdot \dfrac{7}{3} \, hr = \dfrac{7}{2} \, hr = 3 \, hrs \, 30 \, min$

$3 \, hrs \, 30 \, min \, - \, 2 \, hrs \, 20 \, min \, = \, 1 \, hr \, 10 \, min$
 
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