# A person traveled 120 miles in one direction. The return trip was accomplished at double the speed and took 3 hours less the time. How do you find the speed going?

Mar 28, 2016

Velocity going is 20 miles per hour

#### Explanation:

Let distance be $d$

Let velocity going be ${v}_{g}$
Let velocity returning be ${v}_{r}$

Let time going be ${t}_{g}$ hours
Let time returning be ${t}_{r}$ hours

Known: $d = v t \implies v = \frac{d}{t}$ ..................(1)

Given:
$d = 120$ miles
${v}_{r} = 2 {v}_{g}$
${t}_{r} = {t}_{g} - 3$

Objective: determine ${v}_{g}$

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Using equation (1)

${v}_{g} {t}_{g} = d$...................................(2)
${v}_{r} {t}_{r} = d$...................................(3)

But ${t}_{r} = {t}_{g} - 3$ giving

${v}_{g} {t}_{g} = d \text{ } \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . . \left(2\right)$
${v}_{r} \left({t}_{g} - 3\right) = d \text{ } \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . . \left({3}_{a}\right)$

Equate $\left({3}_{a}\right)$ to (2) through $d$

${v}_{g} {t}_{g} = d = {v}_{r} \left({t}_{g} - 3\right)$

But ${v}_{r} = 2 {v}_{g}$ giving

${v}_{g} {t}_{g} = d = 2 {v}_{g} \left({t}_{g} - 3\right)$

${v}_{g} {t}_{g} = 2 {v}_{g} {t}_{g} - 6 {v}_{g}$

But in line with equation (1) $\text{ } {v}_{g} {t}_{g} = 120$ giving

$120 = 2 \left(120\right) - 6 {v}_{g}$

$\implies 6 {v}_{g} = 120$

$\implies {v}_{g} = 20$ miles per hour

Mar 28, 2016

Let's call the speed on the way in ${v}_{1}$ (in mph)
The formula for this is $s = v \times t$

#### Explanation:

Then the speed on the way back ${v}_{2} = 2 \times {v}_{1}$

The time it took the first time is:
$s = {v}_{1} \times {t}_{1} \to {t}_{1} = \frac{s}{v} _ 1 \to {t}_{1} = \frac{120}{v} _ 1$

Riding back:
${t}_{2} = \frac{s}{v} _ 2 = \frac{s}{2 \times {v}_{1}} = \frac{120}{2 \times {v}_{1}} = \frac{60}{v} _ 1$

We know that ${t}_{1} - {t}_{2} = 3$. so:

$\frac{120}{v} _ 1 - \frac{60}{v} _ 1 = 3 \to \frac{60}{v} _ 1 = 3 \to$

$3 \times {v}_{1} = 60 \to {v}_{1} = 20 m p h \to$

${v}_{2} = 2 \times {v}_{1} = 40 m p h$