Question

need some help with this question : When driving to their family reunion, River's mom drove 10 miles at a rate of x mph and then 25 miles at a rate of x + 10 mph . The total driving time was 45 minutes. What were the two driving speeds at which River's mom drove?

Answer 1

To find the two driving speeds, use the distance formula and set up two equations. Simplify the equations to find the speeds.

Step-by-step explanation:

To find the two driving speeds at which River's mom drove, we can use the formula distance = speed x time. We know that the total driving time was 45 minutes. We can set up two equations using the given information:

10 = x ⋅ (45/60)

25 = (x + 10) ⋅ (45/60)

Simplifying the equations, we get:

x = 20 mph

x + 10 = 30 mph

Therefore, River's mom drove at speeds of 20 mph and 30 mph.

Answer 2

The speed of River's mom drove for 10 miles distance is 0.234 mph

The speed of River's mom drove for 25 miles is 10.234 mph

Step-by-step explanation:

Given as :

The first distance cover by River's mom = 10 miles

The rate of speed = x mph

The second distance cover by river's mom = 25 miles

The rate of speed = x + 10 mph

The total driving time = 45 min

Now, According to question

Time =
`(\textrm Distance)/(\textrm speed)`

So, 45 =
`(10)/(x)` +
`(25)/(x + 10)`

Or, 9 =
`(2)/(x)` +
`(5)/(x + 10)`

Or, 9 × x × ( x + 10 ) = 2 × ( x + 10 ) + 5 x

Or, 9 x² + 90 x = 2 x + 20 + 5 x

Or, 9 x² + 90 x = 7 x + 20

Or, 9 x² + 90 x - 7 x - 20 = 0

Or, 9 x² + 83 x - 20 = 0

Solving this quadratic equation

x =
`\frac{-b\pm \sqrt{b^(2)-4* a* c}}{2* a}`

Or, x =
`\frac{-83\pm \sqrt{83^(2)-4* 9* (-20)}}{2* 9}`

Or, x =
`(-83\pm √(7609))/(18)`

∴ x =
`(-83+87.22)/(18)` , tex]\frac{-83-87.22}{18}[/tex]

I.e x = 0.234 , - 9.4567

We consider x = 0.234

So, The speed for 10 miles distance = x = 0.234 mph

and The speed of 25 miles = x + 10 = 0.234 + 10 = 10.234 mph

Hence The speed of River's mom drove for 10 miles distance is 0.234 mph

and The speed of River's mom drove for 25 miles is 10.234 mph Answer