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tigry1 [53]
3 years ago
10

1)Sheyna drive to the lake and back. It took two hours less time to get there than it did to get back. The average speed on the

trip there was 60 mph. The average speed on the way back was 36 mph. How many hours did the trip there take?
Mathematics
1 answer:
ipn [44]3 years ago
8 0

Answer:

8 hours

Step-by-step explanation:

Given:

Sheyna drives to the lake with average speed of 60 mph and

v_1 = 60\ mph

Sheyna drives back from the lake with average speed of 36 mph

v_2 = 36\ mph

It took 2 hours less time to get there than it did to get back.

Let t_1 be the time taken to drive to lake.

Let t_2 be the time taken to drive back from lake.

t_2-t_1 = 2 hrs ..... (1)

To find:

Total time taken = ?

t_1+t_2 = ?

Solution:

Let D be the distance to lake.

Formula for time is given as:

Time =\dfrac{Distance}{Speed }

t_1 = \dfrac{D}{60}\ hrs

t_2 = \dfrac{D}{36}\ hrs

Putting in equation (1):

\dfrac{D}{36}-\dfrac{D}{60} = 2\\\Rightarrow \dfrac{5D-3D}{180} = 2\\\Rightarrow \dfrac{2D}{180} = 2\\\Rightarrow D = 180\ miles

So,

t_1 = \dfrac{180}{60}\ hrs = 3 \ hrs

t_2 = \dfrac{180}{36}\ hrs = 5\ hrs

So, the answer is:

t_1+t_2 = \bold{8\ hrs}

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