You can
put this solution on YOUR website!three key relationships for any vehicle in this case are:
(a) distance = (speed)(time)
(b) speed = distance /time
(c) time = distance /speed
for the car:
distance = d = 15 miles
speed = v
time = t
for the tractor:
distance = D = 10 miles
speed = v - 15
time = T
tractor time = car time, so
T = t
and using key relationship (c), we have:
10/(v-15) = 15/v
we can keep the equivalence if we multiply each side by the product of the denominators
10v = 15(v-15)
that is the same as
10v = 15v - 225
solving for v:
-5v = -225
5v = 225
v = 225/5
v = 45 miles per hour
V = v - 15 = 30 miles per hour
the rate of the tractor is 30 miles per hour
