Question

c. find (limlimits_{t \to 3} \frac{s(t) - s(3)}{t - 3}). show the work that leads to your answer. include units. (s(t) = -4.9t^2 + 508)

Explanation:

Step1: Recall the definition of derivative

The limit $\lim_{t \to a} \frac{s(t) - s(a)}{t - a}$ is the definition of the derivative of $s(t)$ at $t = a$, i.e., $s^\prime(a)$. So we first find the derivative of $s(t)=- 4.9t^{2}+508$ (assuming the original function is $s(t)=-4.9t^{2} + 508$, since the given function in the image seems to have a typo and is likely a quadratic function for a position - time relationship in physics, representing free - fall motion approximately).

The power rule for differentiation is $\frac{d}{dt}(t^{n})=nt^{n - 1}$ and the derivative of a constant $C$ is $0$.

For $s(t)=-4.9t^{2}+508$, the derivative $s^\prime(t)=\frac{d}{dt}(-4.9t^{2})+\frac{d}{dt}(508)$

Step2: Differentiate the function

Using the power rule:

• For the term $-4.9t^{2}$, $n = 2$, so $\frac{d}{dt}(-4.9t^{2})=-4.9\times2t^{2 - 1}=-9.8t$
• For the term $508$ (a constant), $\frac{d}{dt}(508) = 0$

So $s^\prime(t)=-9.8t$

Step3: Evaluate the derivative at $t = 3$

We want to find $\lim_{t\to3}\frac{s(t)-s(3)}{t - 3}=s^\prime(3)$

Substitute $t = 3$ into $s^\prime(t)$:

$s^\prime(3)=-9.8\times3=-29.4$

In the context of a position - time function $s(t)$ (where $s$ is in meters and $t$ is in seconds, typical for free - fall problems), the derivative $s^\prime(t)$ represents velocity. So the units of the limit (which is the velocity at $t = 3$ seconds) are meters per second (m/s).

Answer:

The value of the limit $\lim_{t\to3}\frac{s(t)-s(3)}{t - 3}$ is $- 29.4$ m/s.