Question

A long-distance trucker traveled 120 mi in one direction during a snowstorm. The return trip in rainy weather was accomplished at double the sped and took 3 hr less time. Find the speed going. Using D=rt

Answer 1

Let X be the speed used to travel there.
Time to travel there = Distance / Speed = 120 / X
Thus return trip speed = 2X.
Time to travel back = Distance / Speed = 120 / 2X

Time to travel back = Time to travel there - 3
120 / 2X = 120 / X - 3
120 / 2X = 360 / 3X - 3X / 3X
120 / 2X = 360 - 3X / 3X
120 (3X) = (360 - 3X) (2X)
360X = 720X - 6X^2
6X^2 - 36X = 0
X^2 - 6X = 0
(X)(X-6) = 0
X = ? AND ?

Answer 2

is their anyway you could show me how to do it using the D=RT formula?

Answer 3

D=RT, R is speed, T is time
For travelling there:
120 = RT

For travelling back:
120 = (2R)(T-3)

Answer 4

D=RT, R is speed, T is time
For travelling there:
120 = RT ---(1)

For travelling back:
120 = (2R)(T-3) -----(20

Equate 1 and 2:
RT = (2R)(T-3)
RT = 2RT - 6R
RT - 6R = 0
R(T-6)=0
R=0 or T-6=0
R=0 (N.A), thus T =6

Answer 5

so to find r i would just plug in T into 120=(2r)(T-3)

Answer 6

@dauspex

Answer 7

Yes

Answer 8

and what would be the first original equation before you start adding the properties of the question? Incase i have to do another problem like this

Answer 9

D=RT

This is the general equation.

What i'm doing is to plug in values of D,R,T, if available, then equate them together to try to remove the variables / simplify

Answer 10

ahhh ok, thank you