How much longer does it take in heavy traffic and you drive the same distance at an average speed of only 70km/h

[karush]

211
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.

 

[HallsofIvy]

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.

 

[skeeter]

$r_1t_1 = r_2t_2 \implies t_2 = \dfrac{r_1t_1}{r_2} = \dfrac{(105 \, km/hr)(7/3 \, hr)}{70 \, km/hr} = \dfrac{3}{2} \cdot \dfrac{7}{3} \, hr = \dfrac{7}{2} \, hr = 3 \, hrs \, 30 \, min$

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