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 s(t) = 45t − 5t^2, we need to differentiate the distance function, s(t), with respect to time.
First, let's find the velocity function, v(t), which corresponds to the first derivative of s(t):
v(t) = ds(t) / dt
To find the derivative, we need to differentiate each term of s(t). Remember, when differentiating a power of t, you multiply the original coefficient by the exponent and then subtract 1 from the exponent.
For the first term, 45t, the derivative is 45. Since t has an exponent of 1, the coefficient remains the same.
For the second term, -5t^2, the derivative is -10t. The coefficient, -5, is multiplied by the exponent, 2, resulting in -10t. Afterwards, you subtract 1 from the exponent.
Therefore, the velocity function, v(t), is given by:
v(t) = 45 - 10t
Now, let's find the acceleration function, a(t), which corresponds to the second derivative of s(t):
a(t) = dv(t) / dt
To find the second derivative, we need to differentiate the velocity function, v(t), with respect to time.
For the first term, 45, the derivative is 0 since it does not have a t component.
For the second term, -10t, the derivative is -10.
Therefore, the acceleration function, a(t), is given by:
a(t) = -10
So, the velocity, v(t), is 45 - 10t, and the acceleration, a(t), is -10.