A wrench dropped from a state of rest at time t = 0 travels a distance

$$s(t) = 4.9t^2 m$$

in t seconds. Estimate the instantaneous velocity at t = 3.

A wrench dropped from a state of rest at time $t = 0$ travels a distance $s(t) = 4.9t^2$ m in t seconds. We need to find instantaneous velocity at $t = 3$. Make a table of five average velocity for short interval .average velocity can be find out using the formula $\fbox{Average Velocity $=\dfrac{\Delta s}{\Delta t}$}$

Consider first interval$[t_o, t_1] =[3,3.1]$

$$\begin{align*} \Delta t &=3.1-3.0=0.1\\ \Delta s&=s(3.1)-s(3.0)\\ &=4.9 \times 3.1^2-4.9 \times 3.0^2\\ &=47.089-44.1\\ \Delta s&=2.989\\ Average\ Velocity &=\dfrac{\Delta s}{\Delta t}\\ &=\dfrac{2.989}{0.1}\\ Average\ Velocity &=29.89 \ \ \text{m/s} \end{align*}$$

Given $t=0$ and $s(t)= 4.9t^2$ we will estimate the instantaneous velocity at $t=3$. As discussed in the Textbook we can get the Instantaneous Velocity using the formula :

$$\begin{aligned} \frac{\Delta{s}}{\Delta{t}} \ &= \frac{s(t_1)-s(t_0)}{t_1-t_0}\\ \end{aligned}$$

The instantaneous velocity $(v)$ is given by differentiation w.r.t. to time $t$ of the position function $s(t)$. Given $s(t) = 4.9t^2$ m

$$v(3) = \left . \frac{d}{dt}s(t) \right|_{t=3} = \left . \frac{d}{dt}4.9t^2 \right|_{t=3} = \left . 9.8t \right|_{t=3} = 29.4\ m/s$$

If the ratio between the change in displacement and the change in time is observed, the definition of the average velocity is obtained. Mark average velocity with $\color{#c34632}s_a$.

If the time interval $(t_1, t_2)$ is observed, the average velocity is calculated as:

$$\color{#c34632}{s_a}\color{black}{=}\color{black}\frac{s(t_2)-s(t_1)}{t_2-t_1}=\frac{\Delta s}{\Delta t}.$$