Question

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?

Answer 1

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:

$t_1+t_2 = \bold{8\ hrs}$