Answer:
8 hours
Step-by-step explanation:
Given:
Sheyna drives to the lake with average speed of 60 mph and
[tex]v_1 = 60\ mph[/tex]
Sheyna drives back from the lake with average speed of 36 mph
[tex]v_2 = 36\ mph[/tex]
It took 2 hours less time to get there than it did to get back.
Let [tex]t_1[/tex] be the time taken to drive to lake.
Let [tex]t_2[/tex] be the time taken to drive back from lake.
[tex]t_2-t_1 = 2[/tex] hrs ..... (1)
To find:
Total time taken = ?
[tex]t_1+t_2 = ?[/tex]
Solution:
Let D be the distance to lake.
Formula for time is given as:
[tex]Time =\dfrac{Distance}{Speed }[/tex]
[tex]t_1 = \dfrac{D}{60}\ hrs[/tex]
[tex]t_2 = \dfrac{D}{36}\ hrs[/tex]
Putting in equation (1):
[tex]\dfrac{D}{36}-\dfrac{D}{60} = 2\\\Rightarrow \dfrac{5D-3D}{180} = 2\\\Rightarrow \dfrac{2D}{180} = 2\\\Rightarrow D = 180\ miles[/tex]
So,
[tex]t_1 = \dfrac{180}{60}\ hrs = 3 \ hrs[/tex]
[tex]t_2 = \dfrac{180}{36}\ hrs = 5\ hrs[/tex]
So, the answer is:
[tex]t_1+t_2 = \bold{8\ hrs}[/tex]
