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?

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Answer 1 · 2016-03-28T11:40:35

Velocity going is 20 miles per hour

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 · 2016-03-28T11:47:15

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

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#