rate time distance algebra problem very hard!?
On October 29, Dena drove 222.7 miles at a constant speed. One week later (November 5) she drove the same 222.7 miles at a constant speed 5 mph faster than the previous week and reduced the driving time by 10 minutes. How fast did Dena drive on November 5 and how long did it take her to drive 222.7 miles on that date?
i keep getting a negative number
ok so guess and check reveals that the speeds are 111.9mph and 116.9 mph (i think) but how the heck do i set up an equation
Comments
222.7 = v * t
222.7 = (v + 5) * (t - 1/6)
(v + 5) * (t - 1/6) = vt
vt + 5t - v/6 - 5/6 = vt
5t = (v + 5) / 6
30t = v + 5
222.7 = (v + 5) * (t - 1/6)
222.7 = (30t) * (t - 1/6)
222.7 = 30t^2 - 5t
0 = 30t^2 - 5t - 222.7
t = (5 +/- sqrt(25 + 4 * 30 * 222.7)) / 60
t = (5 +/- sqrt(25 + 12 * 2227)) / 60
t = (5 +/- 163.55121522018722071712395359725) / 60
t = 168.55121522018722071712395359725 / 60 (we can discard the 5 - .... answer since that will be negative, and have you ever heard of negative time?)
t = 2.8091869203364536786187325599542
222.7 = v * t
222.7 = v * 2.8091869203364536786187325599542
222.7 / 2.8091869203364536786187325599542 = v
v = 79.275607610093610358561976798626
So, she drove 84.275 mph on Nov 5, and it took her 2.809 hours to make the trip
D = RT
On Oct 29,
D = 222.7
R = R,
T = 222.7/R
On Nov 5
D = 222.7
R = R+5
T = 222.7/(R+5) = 222.7/R - 1/6 (Because 10 min = 1/6 hr)
222.7/(R+5) = 222.7/R - 1/6
1/(R+5) - 1/R = -1/(6*222.7)
R - (R+5) = -R(R+5)/1336.2
5*1336.2 = R^2 + 5R
R^2 + 5R - 6681 = 0
R = 79.2756 (using the quadratic formula)
On Nov 5
R = 79.2756 + 5 = 84.2756 mph
T = 222.7/84.2756 = 2.64252049 hours
This is a set of equations You know that distance is a constant and that the speeds are relative and times are relative.
Relative Speed
v1 + 5 = v2
note that time is in minutes and should be switched to hours to match rate, so 10 minutes is equal to 1/6 of an hour.so
t1 - 1/6 = t2
Write the distance equation
222.7 = v1 * t1 = v2 * t2
substitute and solve.
t is the time it takes for the 2nd practice moving at sixty 8 km/h to seize up so it incredibly is organic write the 1st equation the place t is the time the 2nd practice traveled until now assembly the 1st practice. in the 2nd equation, t is the time that the 1st practice traveled until now the 2nd practice caught up.
d=rt