Jenny had driven for 2 h at a constant speed when road repairs forced her to reduce her speed by 10 mi/h for the remaining 1 h of her 152-mile trip. Find her original speed.
2007-11-16 07:08:35 · 6 answers · asked by dani m 1 in Science & Mathematics ➔ Mathematics
Apply rate * time = distance to both segments of the trip, then add the distances
V * 2 = D1
(V - 10) * 1 = D2
D1 + D2 = 152
So
V * 2 + (V-10) * 1 = 152
Solve for V
2007-11-16 07:13:44 · answer #1 · answered by jgoulden 7 · 1⤊ 0⤋
let the original speed = x mi/h
Distance travelled at x mi/h for 2 h = 2*x = 2x mi
after reduction, the speed = (x - 10) mi/h
distance travelled at (x-10) mi/h for 1 h = (x-10)*1 = (x-10) mi
total distance travelled = 2x + x - 10 = 3x - 10
so 3x - 10 = 152
3x = 162
x = 54 mi/h
the original speed = 54 mi/h
2007-11-16 15:29:03 · answer #2 · answered by mohanrao d 7 · 0⤊ 0⤋
using d = r * t :
r1,t1 are the speed and time of the first part of the trip
r2, t2 are the speed and time of the second part
r1 = x then r2 = x - 10 , from the info given.
d = r1*t1 + r2*t2
152 = 2x + x - 10
x = 54 mph = r1 = original speed
2007-11-16 15:19:17 · answer #3 · answered by KEYNARDO 5 · 0⤊ 0⤋
her original speed would be 54 mph
2007-11-16 15:21:24 · answer #4 · answered by Anonymous · 0⤊ 0⤋
54 mph
2007-11-16 15:13:30 · answer #5 · answered by Matt 3 · 0⤊ 0⤋
2V+(V-10)=152
Add the 'V's' together
3V-10=152
Add 10 to both sides
3V=162
Divide by 3
v=54mph
2007-11-16 15:21:56 · answer #6 · answered by bostep662 4 · 0⤊ 0⤋