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
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
Well, let's see if we can figure this out. We're given three points on the graph of the position equation: (1, 161), (2, 98), and (3, 3).
We'll start with the general form of the position equation: s = a * t^2 + b * t + c. We need to find the values of a, b, and c.
First, let's plug in the values from the first point, (1, 161):
161 = a * 1^2 + b * 1 + c
161 = a + b + c (Eq. 1)
Next, let's plug in the values from the second point, (2, 98):
98 = a * 2^2 + b * 2 + c
98 = 4a + 2b + c (Eq. 2)
Finally, let's plug in the values from the third point, (3, 3):
3 = a * 3^2 + b * 3 + c
3 = 9a + 3b + c (Eq. 3)
Now, we have a system of three equations with three unknowns (a, b, and c). We can solve this system to find the values of a, b, and c.
Solving Eq. 1 and Eq. 2, we get:
161 - 98 = a + b + c - (4a + 2b + c)
63 = -3a - b
Solving Eq. 2 and Eq. 3, we get:
98 - 3 = 4a + 2b + c - (9a + 3b + c)
95 = -5a - b
Now we have a system of two equations with two unknowns (a and b). Solving this system, we find:
-3a - b = 63 (Eq. 4)
-5a - b = 95 (Eq. 5)
Adding Eq. 4 and Eq. 5, we get:
-8a - 2b = 158
Simplifying, we find:
8a + 2b = -158 (Eq. 6)
Multiplying Eq. 4 by 2, we get:
-6a - 2b = 126 (Eq. 7)
Adding Eq. 6 and Eq. 7, we get:
2a = -32
Dividing by 2, we find:
a = -16
Now, we can substitute the value of a back into Eq. 4 to solve for b:
-3(-16) - b = 63
48 - b = 63
b = -15
Finally, substituting the values of a and b back into Eq. 1, we can solve for c:
161 = -16 * 1^2 - 15 * 1 + c
161 = -16 - 15 + c
161 = -31 + c
c = 192
So, the position equation for the object is:
s = -16t^2 - 15t + 192
Therefore, the answer is (e) s = -16t^2 - 15t + 192.
To find the position equation, we first need to determine the values of the coefficients in the equation.
Given:
At t = 1 second, s = 161 feet (Equation 1)
At t = 2 seconds, s = 98 feet (Equation 2)
At t = 3 seconds, s = 3 feet (Equation 3)
We can substitute these values into the position equation, s = 12at^2 + vot + so, and solve for the coefficients a, vo, and so.
1. Substitute t = 1 and s = 161 into the position equation:
161 = 12a(1^2) + vo(1) + so
161 = 12a + vo + so ...(Equation 4)
2. Substitute t = 2 and s = 98 into the position equation:
98 = 12a(2^2) + vo(2) + so
98 = 48a + 2vo + so ...(Equation 5)
3. Substitute t = 3 and s = 3 into the position equation:
3 = 12a(3^2) + vo(3) + so
3 = 108a + 3vo + so ...(Equation 6)
Now, we have three equations (Equations 4, 5, and 6) with three unknowns (a, vo, and so). We can solve these equations simultaneously to find the values of the coefficients.
Solving equations 4, 5, and 6 simultaneously:
Equation 5 - Equation 4:
98 - 161 = 48a + 2vo + so - (12a + vo + so)
-63 = 36a + vo ...(Equation 7)
Equation 6 - Equation 5:
3 - 98 = 108a + 3vo + so - (48a + 2vo + so)
-95 = 60a + vo ...(Equation 8)
Now we have a system of equations: Equation 7 and Equation 8.
Adding Equation 7 and Equation 8:
-63 + (-95) = 36a + vo + 60a + vo
-158 = 96a + 2vo ...(Equation 9)
Equation 8 - Equation 7:
-95 - (-63) = 60a + vo - 36a - vo
-32 = 24a ...(Equation 10)
Now, we can solve Equation 9 and Equation 10.
From Equation 10:
-32 = 24a
a = -32 / 24
a = -4/3
Substituting the value of a into Equation 7:
-158 = 96(-4/3) + 2vo
-158 = -128 + 2vo
2vo = -158 + 128
2vo = -30
vo = -30 / 2
vo = -15
Now, we have found the values of a = -4/3 and vo = -15.
Substituting these values into Equation 4:
161 = 12(-4/3)(1^2) + (-15)(1) + so
161 = -16 + (-15) + so
161 = -31 + so
so = 161 + 31
so = 192
Therefore, the position equation for the object is:
s = -16t^2 - 15t + 192
The correct answer is (e) s=−16t^2 −15t+192.
To find the position equation for the object, we can use the given heights at different times to form a system of equations.
Given:
At t = 1 second, s = 161 feet
At t = 2 seconds, s = 98 feet
At t = 3 seconds, s = 3 feet
Let's plug these values into the position equation s = 12t^2 + v₀t + s₀ and solve for the unknowns.
For t = 1:
161 = 12(1)^2 + v₀(1) + s₀
161 = 12 + v₀ + s₀ --- Equation 1
For t = 2:
98 = 12(2)^2 + v₀(2) + s₀
98 = 48 + 2v₀ + s₀ --- Equation 2
For t = 3:
3 = 12(3)^2 + v₀(3) + s₀
3 = 108 + 3v₀ + s₀ --- Equation 3
Now we have a system of three equations with two unknowns (v₀ and s₀). We can solve this system using various methods, such as substitution or elimination.
Let's solve it by elimination:
First, subtract Equation 1 from Equation 2:
98 - 161 = 48 + 2v₀ + s₀ - (12 + v₀ + s₀)
-63 = 36 + v₀ --- Equation 4
Next, subtract Equation 1 from Equation 3:
3 - 161 = 108 + 3v₀ + s₀ - (12 + v₀ + s₀)
-158 = 96 + 2v₀ --- Equation 5
Now we have a system of two equations:
-63 = 36 + v₀ --- Equation 4
-158 = 96 + 2v₀ --- Equation 5
Solving Equation 4:
v₀ = -63 -36
v₀ = -99
Substitute the value of v₀ into Equation 5:
-158 = 96 + 2(-99)
-158 = 96 - 198
-158 = -102
As Equation 5 is not satisfied, there is no consistent solution for the system of equations. This means the given heights at different times do not correspond to a valid position equation.
Therefore, none of the answer choices (a, b, c, d, e) are correct.