An object has a constant acceleration of 40 ft/sec2, an initial velocity of -20 ft/sec, and an initial position of 10 ft. Find the position function, s(t), describing the motion of the object.
d^2s/dt^2 = a = 40
ds/dt = v = 40 t - 20 [so v is -20 when t = 0]
s = (40/2 ) t^2 - 20 t + 10 [ so s is 10 when t = 0)
s = 20 t^2 - 20 t + 10 feet
To find the position function, we need to integrate the acceleration function twice with respect to time.
Given:
Acceleration (a) = 40 ft/sec^2
Initial velocity (v0) = -20 ft/sec
Initial position (s0) = 10 ft
First, let's find the velocity function, v(t):
v(t) = v0 + ∫(a dt)
v(t) = -20 + ∫(40 dt) [since acceleration is constant]
v(t) = -20 + 40t + C1 [integrating with respect to time, where C1 is the constant of integration]
Next, let's find the position function, s(t):
s(t) = s0 + ∫(v dt)
s(t) = 10 + ∫((-20 + 40t + C1) dt) [substituting the velocity function]
s(t) = 10 + (-20t + 20t^2/2 + C1t + C2) [integrating with respect to time, where C2 is the constant of integration]
Simplifying further:
s(t) = -20t + 10t^2 + C1t + C2
In order to determine the constants C1 and C2, we can use the initial conditions:
When t = 0, s(0) = 10 ft
So, s(0) = -20(0) + 10(0)^2 + C1(0) + C2
10 = C2
When t = 0, v(0) = -20 ft/sec
So, v(0) = -20 + 40(0) + C1
-20 = C1
Thus, the position function is:
s(t) = -20t + 10t^2 - 20t + 10
s(t) = 10t^2 - 40t + 10
Therefore, the position function s(t) describing the motion of the object is 10t^2 - 40t + 10 ft.