A band's equipment truck travelled a distance of 720 km. On the return trip, the average speed was increased by 10 km/h. If the total driving time for the round trip was 17 h, what was the average speed?
2007-03-21 17:55:07 · 5 answers · asked by john 1 in Science & Mathematics ➔ Mathematics
Let x km/h be the speed
and (x+10) km/k speed on the return trip
Time=distance/speed
total time for the round trip is 17 hours
(720/x)+(720/(x+10))=17
solve the equation(multiply both sides by
x(x+10) to eliminate the denominators)
720(x+10)+720x=17x(x+10) , and you get a quadratic equation:
17x^2-1270x-7200=0
solve using quadratic formula:
x1=80
x2=-90/17 reject (it is a negative, and x is average speed which must be >0)
Answer: 80km/h
expand
2007-03-24 16:02:59 · answer #1 · answered by Joanna 1 · 0⤊ 0⤋
The average speed would be
total distance/total time=720/17= 42.35 kmph
2007-03-21 18:32:32 · answer #2 · answered by Chris 521n36 1 · 0⤊ 0⤋
this looked similar to the cottage one. one speed going to and another diff speed coming from. thinking that way I figured 80 km/h there and 90 km/h coming back which would be right but you wanted avg speed. Total Distance/Total time. 720/17 42.35
2007-03-21 22:45:54 · answer #3 · answered by niceguyswthrt 2 · 0⤊ 0⤋
y= speed going out
y+10 = speed returning
y + 5 = average speed
distance travelled/rate = time
720/(y+5) = 17 hours
720 = 17y + 85
17y = 635
y = 37.35
average speed = 42.35
(sorry for all the redos)
2007-03-21 17:59:54 · answer #4 · answered by Jim S 5 · 0⤊ 0⤋
720=s1*t1=>t1=720/s1
720=s2*t2=(s1+10)t2
s2=s1+10
t1+t2=17 h=>t2=17-t1
720=(s1+10)t2=(s1+10)(17-t1)=(s1+10)(17-720/s1)
720=17s1+170-720-7200/s1=>
17s1-7200/s1-1270=0
17s1^2-1270s1-7200=0
s1=(1270+1450)/34=80
s2=90
1440/17=84.70 average speed
2007-03-21 18:36:01 · answer #5 · answered by djin 2 · 0⤊ 0⤋