Question

the position (in meters) at time t (in seconds) of a particle that moves along a straight line is given by the function s(t). the first derivative of s(t) is called the velocity, denoted by v(t); that is, the velocity is the rate of change of the position. the rate of change of the velocity is called acceleration, denoted by a(t); that is, \\( \frac{d}{dt}v(t)=a(t) \\).
given that \\( v(t)=s(t) \\), it follows that \\( \frac{d^{2}}{dt^{2}}s(t)=a(t) \\).
find the velocity a d the acceleration at time \\( t = 5 \\) s for the position function \\( s(t)=t^{2}-4t \\).
\\( v(5)= \\)

Explanation:

Step1: Find the velocity function

Differentiate the position function \(s(t)=t^{2}-4t\) with respect to \(t\).
Using the power rule \(\frac{d}{dt}(t^{n}) = nt^{n - 1}\), we have \(v(t)=\frac{d}{dt}(t^{2}-4t)=2t - 4\).

Step2: Calculate the velocity at \(t = 5\)

Substitute \(t = 5\) into the velocity function \(v(t)\).
\(v(5)=2\times5-4\).

$$
LATEXBLOCK0
$$

Step3: Find the acceleration function

Differentiate the velocity function \(v(t)=2t - 4\) with respect to \(t\).
Using the power rule, \(a(t)=\frac{d}{dt}(2t - 4)=2\).
Since the acceleration function \(a(t)\) is a constant function, \(a(5)=2\).

Answer:

The velocity \(v(5) = 6\) m/s and the acceleration \(a(5)=2\) m/s².