hours. what was his speed each way.
2007-08-13 08:01:51 · 5 answers · asked by sugar lady 1 in Science & Mathematics ➔ Mathematics
His speeds were 40 mph out, and 30 mph back.
Here's how this was found. Let his outward speed be v mph, then the distance driven outwards was 3v. Coming back his speed was (v - 10) mph, and the distance 4(v - 10).
So 3v = 4(v - 10), that is v = 40.
QED
(CHECK: both sides are 3*40 = 4*30 = 120, so it's correct.)
Live long and prosper.
2007-08-13 08:08:16 · answer #1 · answered by Dr Spock 6 · 0⤊ 0⤋
Formula is d=rt (distance=ratextime) . So if the distances are equal, then the ratextime would be equal. The rate in #1 is unknown, but the time is 3 hours. The rate in #2 is (r-10), and the time is 4. The equation would be rt=rt, or (r)(3)=(r-10)(4). 3r=(r-10)4: 3r=4r-40: solve for r; r=40, so, r-10=30
Check: 3r=3x40=120 and (r-10)4=(30x4) =120
2007-08-13 15:22:27 · answer #2 · answered by Ed S 4 · 0⤊ 0⤋
Let s = first speed
s-10 = 2nd speed
3·s = 4 (s-10)
3s = 4s-40
0 = s-40
s = 40 mph
s - 10 = 30 mph
2007-08-13 15:11:44 · answer #3 · answered by Tony The Dad 3 · 0⤊ 0⤋
distance = speed*time
1st trip
distance = 3*v
2nd
distance = 4*(v-10)
so
3*v = 4*v -40
so
v = 40
so 1st trip speed = 40mph
2nd trip speed = 30mph
and the distance is 120 miles
2007-08-13 15:12:48 · answer #4 · answered by Mike 5 · 0⤊ 0⤋
going, d/r = 3
return, d/(r-10) = 4
3r = d = 4(r - 10)
3r = 4r - 40
r (going) = 40 mph
return, r-10 = 30 mph
2007-08-13 15:12:34 · answer #5 · answered by Philo 7 · 0⤊ 0⤋