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)
To determine the velocity and acceleration as functions of time for the given position function s(t) = 45t − 5t^2, we need to take the first and second derivatives of s(t) with respect to time.
1. Find the velocity function:
To find the velocity function, we differentiate s(t) with respect to time (t).
s'(t) = d/dt (45t − 5t^2)
To differentiate, we use the power rule and constant rule for differentiation. The power rule states that if y = a*x^n, then dy/dx = n*a*x^(n-1).
Using the power rule:
s'(t) = 45 * d/dt(t) − 5 * d/dt(t^2)
s'(t) = 45 * (1) - 5 * 2t
s'(t) = 45 - 10t
Therefore, the velocity function is v(t) = 45 - 10t.
2. Find the acceleration function:
To find the acceleration function, we differentiate v(t) with respect to time (t).
v'(t) = d/dt(45 - 10t)
Using the constant rule:
v'(t) = -10 * d/dt(t)
v'(t) = -10 * (1)
v'(t) = -10
Therefore, the acceleration function is a(t) = -10.
The velocity as a function of time is v(t) = 45 - 10t, and the acceleration as a function of time is a(t) = -10.