subbing in your values, you will end up with 3 equations, in 3 unknowns
#1.
161 = 12a + v + c , I will use c to avoid confusion with a second s
#2.
98 = 48a + 2v + c
#3.
3 = 108a + 3v + c
#3 - #2 ------> 60a + v = -95
#2 - #1 ------> 36a + v = -63
subtract those two ...
24a = -32
a = -32/24 = - 4/3
into 36a + v = -63
-48 + v = -63
v = -15
into #1:
12(-4/3) - 15 + c = 161
c = 192
s = 12(-4/3)t^2 - 15t + 192
= -16t^2 - 15t + 192
An object moving vertically is at the given heights at the specified times. Find the position equation s = 12 at^2 + v (small o) t + s (small o) for the object.
At t = 1 second, s = 161 feet At t = 2 seconds, s = 98 feet At t = 3 seconds, s = 3 feet
a. s=−8t^2−t−192
b. s=−32t^2 −15t+161
c. s=−16t2 +15t+161
d. s=−16t^2 −15t−192
e. s=−16t2 −15t+192
1 answer
1 answer