A golfer rides in a golf cart at a speed of 3.10 m/s for 27.0 s. She then gets out of the cart and starts walking at an average speed of 1.30 m/s. For how long (in seconds) must she walk if her average speed for the entire trip, riding and walking, is 2.20 m/s?
2006-09-25 07:19:07 · 2 answers · asked by Alan l 1 in Education & Reference ➔ Homework Help
Average speed = (speed riding * time riding + speed walking * time walking )/total time
2.20 m/s = (3.10 m/s * 27 sec + 1.3 m/s * x) / (x+27) where x is the time spent walking
2.20 m/s = (83.7 + 1.3x) / (x+27)
2.20 (x+27) = (83.7 + 1.3x)
2.2x + 59.4 = 83.7 + 1.3x
.9x = 24.3
x = 27 sec (solution)
Check:
2.20 m/s = (3.10 m/s * 27 sec + 1.3 m/s * x) / (x+27)
2.20 m/s = (3.10 m/s * 27 sec + 1.3 m/s * 27) / (27+27)
2.2 = (83.7 + 35.1) / 54
2.2 = 118.8 / 54
2.2 = 2.2 (check!)
2006-09-25 07:26:52 · answer #1 · answered by ³√carthagebrujah 6 · 0⤊ 0⤋
Speed = distance/time
Distance:
3.1m/s x 27s + 1.3m/s x t
83.7m + 1.3t
Time:
27+t
Speed:
(83.7+1.3t)/(27+t) = 2.2
83.7+1.3t = 2.2(27+t)
83.7+1.3t = 59.4+2.2t
83.7-59.4=2.2t-1.3t
24.3 = 0.9t
t=27s
2006-09-25 14:31:34 · answer #2 · answered by Jenelle 3 · 0⤊ 0⤋