An object moving vertically is at the given heights at the specified times. Find the position equation
s = at2 + v0t + s0 for the object given the following conditions.
• At t = 1 seconds, s = 48 feet
• At t = 2 seconds, s = 64 feet
• At t = 3 seconds, s = 48 feet
To find the position equation for the object, we need to determine the values of the acceleration (a), initial velocity (v0), and initial position (s0) in the given equation s = at^2 + v0t + s0.
Given the information, we can form a system of equations using the given conditions:
Equation 1: 48 = (1^2)a + (1)v0 + s0
Equation 2: 64 = (2^2)a + (2)v0 + s0
Equation 3: 48 = (3^2)a + (3)v0 + s0
We can solve this system of equations to find the unknown values of a, v0, and s0.
To do this, we can use a method such as substitution or elimination.
Using the substitution method:
From Equation 1, we can express s0 in terms of a and v0:
s0 = 48 - a - v0
Substituting this expression in Equations 2 and 3, we have:
Equation 2: 64 = (2^2)a + (2)v0 + (48 - a - v0)
Equation 3: 48 = (3^2)a + (3)v0 + (48 - a - v0)
Simplifying both equations, we get:
4a + 2v0 + 48 - a - v0 = 64
9a + 3v0 + 48 - a - v0 = 48
Combining like terms, we have:
3a + v0 = 16 (Equation 4)
8a + 2v0 = 0 (Equation 5)
From Equation 5, we can express v0 in terms of a:
v0 = -4a
Substituting this expression in Equation 4:
3a - 4a = 16
-a = 16
Dividing both sides by -1, we get:
a = -16
Now that we have the value of acceleration, we can substitute it back into Equation 4 to find v0:
3(-16) + v0 = 16
-48 + v0 = 16
v0 = 16 + 48
v0 = 64
Finally, we can substitute the values of a and v0 into Equation 1 to find s0:
48 = (1^2)(-16) + (1)(64) + s0
48 = -16 + 64 + s0
48 = 48 + s0
s0 = 0
Therefore, the position equation for the object is:
s = -16t^2 + 64t + 0
s = -16t^2 + 64t