Question

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?

Answer 1

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=vt => v=d/t` ..................(1)

Given:
`d=120` miles
`v_r=2v_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" "...................................(2)`
`v_r (t_g-3)=d" "..........................(3_a)`

Equate `(3_a)` to (2) through `d`

`v_g t_g = d=v_r (t_g-3)`

But `v_r=2v_g` giving

`v_g t_g = d=2 v_(g) (t_g - 3)`

`v_g t_g=2 v_g t_g-6v_g`

But in line with equation (1) `" "v_g t_g=120` giving

`120=2(120)-6v_g`

`=>6v_g=120`

`=>v_g=20` miles per hour

Answer 2

Let's call the speed on the way in `v_1` (in mph)
The formula for this is `s=vxxt`

Explanation:

Then the speed on the way back `v_2=2xxv_1`

The time it took the first time is:
`s=v_1xxt_1->t_1=s/v_1->t_1=120/v_1`

Riding back:
`t_2=s/v_2=s/(2xxv_1)=120/(2xxv_1)=60/v_1`

We know that `t_1-t_2=3`. so:

`120/v_1- 60/v_1=3->60/v_1=3->`

`3xxv_1=60->v_1=20mph->`

`v_2=2xxv_1=40mph`