I need help please!

Question

As part of Jane's workout, she runs to the store, buys a snack, and then walls home. She spends ten minutes in the store, and the whole trip takes one hour. Her running speed is 6 mph and her walking speed is 3 mph. How far is she from the store?

Answer 1

speed/distance/time equation s = d/t (or t = d/s or d = st)...

x mph = x/60 mpmin (e.g. 6 mph = 0.1 mpmin)

60 minutes total

-10 minutes in store = 50 minutes

running speed = 2x walking speed, and same distance there & back, so 2/3 time walking (2/3 x 50) and 1/3 time running (1/3 x 50)

d = st = 6/60 x 50/3 = 1/10 x 50/3 = 1 x 5/3 = 1 ²/₃ miles from store

(or d = st = 3/60 x 100/3 = 1/20 x 100/3 = 1 x 5/3 = 1 2/3 miles from store)

to check... 5/3 miles from store... run at 6mph = 5/18 of hour

5/3 miles from store... walk at 3mph = 5/9 of hour

5/9 + 5/18 = 15/18 hour

15/18 x 60 = 50 minutes

Answer 2

Another way to look at it:

Let x = time to run to the store and y = time to walk home

x + y = (60-10) mins/(60 mins/hour) = 50/60 = 5/6 hour (Equation 1)

Distance is velocity • time or vt. The distance both ways is the same so:

distance = 6x = 3y or y = 2x

Substituting the value for y into Equation 1

x + 2x = 5/6 --> 3x = 5/6 -> x = 5/18 hours and y = 2x = 5/9 hours (not really needed)

distance = velocity * time = 6(5/18) = 5/3 ≈ 1.67 miles. Also 3(5/9) = 5/3.

Answer 3

Let's set up the equation for time. We know velocity equals distance divided by time or

v = d/t

so we can solve for t by saying,

t = d/v

We need 3 different parts for time:

Running to the store: t = d/6

Spending time in the store: t = 10 minutes = 0.17 hours

Walking home from the store: t = d/3

Since we know the whole trip takes 1 hour, we can set up the following equation:

d/6 + 0.17 + d/3 = 1

d/6 + 2d/6 = 0.83

3d/6 = 0.83

d/2 = 0.83

d = 1.67 miles

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