An object has a constant acceleration of 30 ft/sec2, an initial velocity of −10 ft/sec, and an initial position of 4 ft. Find the position function, s(t), describing the motion of the object.
To find the position function, we need to integrate the acceleration function with respect to time twice.
Let's start by integrating the acceleration function. The acceleration function is given as a constant 30 ft/sec^2. When we integrate a constant acceleration, we get the velocity function.
The integral of a constant is equal to the constant multiplied by the variable we are integrating with respect to. In this case, we integrate with respect to time.
∫(30) dt = 30t + C1
Here, C1 represents the constant of integration. Since the initial velocity is given as -10 ft/sec, we can find the value of C1 by substituting the initial condition.
-10 = 30(0) + C1
C1 = -10
Therefore, the velocity function v(t) is:
v(t) = 30t - 10
Now, let's integrate the velocity function to find the position function.
∫(30t - 10) dt = 15t^2 - 10t + C2
Again, C2 represents the constant of integration. We can find the value of C2 by substituting the initial position condition.
4 = 15(0)^2 - 10(0) + C2
C2 = 4
Therefore, the position function s(t) is:
s(t) = 15t^2 - 10t + 4
So, the position function describing the motion of the object is s(t) = 15t^2 - 10t + 4.
just review your text, and you will see why
s(t) = 4 -10t + 15t^2