Question 158862
1. two cars enter a Florida Turnpike at 8:00am, each heading for Wildwood. one car average speed is 10 miles per hour more than the other. The faster car arrives at Wildwood at 11:00am, 1/2 hours before the other car. What is the average speed of each car? How far did each travel.
:
Fast car travel time = 3 hrs
Slow car travel time = 3.5 hrs
:
Let s = speed of the slower car
then
(x+10) = speed of the faster car
:
Both cars traveled the same distance, write a distance equation
Dist = time * speed
:
slow car dist = fast car dist
3.5s = 3(s+10)
3.5s = 3s + 30
3.5s - 3s = 30
.5s = 30
s = 60 mph is the slow car
and
60 + 10 = 70 mph is the speed of the fast car
:
Find the dist using the slow car expression
3.5 * 60 = 210 mi
Check solutions by finding the dist using the fast car expression
3 * 70 = 210 mi
:
:
2. Am object is propelled upward with an initial velocity of 20 meters per second. The distance, in meter, of the object from the ground after t seconds is s=-4.9t+20t. I think this should be -4.9t^2 + 20t
:
a. when will the object be 15 meters above the ground?
:
-4.9t^2 + 20t = 15
-4.9t^2 + 20t - 15 = 0
Use the quadratic formula, we get two positive solutions:
t ~ 1 sec (on the way up)
and
t ~ 3.1 (on the way down)

b. when will it strike the ground?
:
When it strikes the ground, s = 0
-4.9t^2 + 20t = 0
-t(4.9t - 20) = 0
4.9t - 20 = 0
4.9t = +20
t = {{{20/4.9}}}
t = 4.1 sec
;
:
c. will the object reach a height of 100 meters?
;
Obviously it won't, max will occur around 2 sec
h = -4.9(2^2) + 20(2)
h = -19.6 + 40 = 20.4 ft, not even close