2. The amount of gasoline used by a car
varies jointly as the distance travelled
and the square root of the speed
Suppose a car used 25 liters on a 100
km-trip at 100 kph, about how many
liters will it use on a 1000 km trip at 64
kph?
A. 100 L
B. 200 L
C. 300 L
D. 400 L
JOINT VARIATION
==============================
» Assign variables to each following:
- Let (g) be the amount of gasoline.
- Let (d) be the distance travelled.
- Let (s) be the speed.
[tex] \: : \implies \sf \large g = kd \sqrt{s} [/tex]
[tex]\large \tt \red{given} \begin{cases} \sf \: g = 25 \\ \sf \: d = 100 \\ \sf \: s = 100\end{cases}[/tex]
» Substitute to find the constant (k).
[tex]\implies \sf \large 25 = k(100) (\sqrt{100})[/tex]
[tex]\implies \sf \large 25 = k(100)(10)[/tex]
[tex]\implies \sf \large 25 = k(1000)[/tex]
[tex]\implies \sf \large \frac{25}{1000} = \frac{k( \cancel{1000})}{ \cancel{1000}} \\ [/tex]
[tex]\implies \sf \large \frac{1}{40} =k \\ [/tex]
[tex]\implies \sf \large k = \frac{1}{40} \\ [/tex]
» Now find how many liters (g) will be used if the distance (d) is 1000 km and the speed (s) is 64 kmph.
[tex]\large \tt \red{given} \begin{cases} \sf \: k = \frac{1}{40} \\ \sf \: d = 1000 \\ \sf \: s = 64\end{cases}[/tex]
[tex] \implies \sf \large g = \frac{1}{40} (1000)( \sqrt{64} )[/tex]
[tex] \implies \sf \large g = \frac{1}{40} (1000)( 8 )[/tex]
[tex] \implies \sf \large g = \frac{1}{40} (8000)[/tex]
[tex] \implies \sf \large g = \frac{8000}{40} \\ [/tex]
[tex] \implies \sf \large g = 200 \\ \\ [/tex]
Final Answer:
[tex] \tt \huge» \: \purple{B. \: 200\, L }[/tex]
==============================
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