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.
s(t)=4-10t+15^2
looks good to me.
thanks!
The position function, s(t), describes the position of an object as a function of time. To find the position function for the given scenario, we need to integrate the acceleration function twice with respect to time.
Given:
Constant acceleration = 30 ft/sec^2
Initial velocity = -10 ft/sec
Initial position = 4 ft
First, let's find the first antiderivative of the acceleration function to find the velocity function. The integral of the constant acceleration, 30 ft/sec^2, with respect to time is 30t. However, since we have an initial velocity of -10 ft/sec, we need to add this to the result. So the velocity function, v(t), is given by:
v(t) = 30t - 10
Next, we integrate the velocity function to find the position function. The integral of (30t - 10) with respect to time is (15t^2 - 10t). Again, we need to consider the initial position of 4 ft, so the position function, s(t), is given by:
s(t) = 15t^2 - 10t + 4
Therefore, the position function, s(t), describing the motion of the object is:
s(t) = 15t^2 - 10t + 4