Determine the velocity and acceleration as functions of time, t, for s(t) = 45t − 5t^2�, where
s(t) represents the distance as a function of time. (Hint: velocity and acceleration correspond to the first and
second derivatives of the distance)
v = 45 - 10t
a = -10
To determine the velocity and acceleration as functions of time for the given distance function, we need to take the first and second derivatives of the distance function, s(t).
Let's start with finding the first derivative, which represents the velocity.
1. Take the derivative of each term in the distance function, s(t).
s(t) = 45t − 5t^2
ds/dt = d(45t)/dt - d(5t^2)/dt
ds/dt = 45 - 10t
Therefore, the velocity function, v(t), is given by ds/dt = 45 - 10t.
Now, let's find the second derivative, which represents the acceleration.
2. Take the derivative of the velocity function, v(t).
dv/dt = d(45 - 10t)/dt
dv/dt = -10
Therefore, the acceleration function, a(t), is a constant, -10.
To summarize:
- The velocity function, v(t), is given by 45 - 10t.
- The acceleration function, a(t), is a constant, -10.
These functions can be used to analyze the motion of the object represented by the distance function, s(t).