Category: Open Textbook Library

Posted on

202110091141 Exercises 1.1.A (Q1-Q4)

For Exercises 1-4, suppose that an object moves in a straight line such that its position s after time t is the given function s=s(t). Find the instantaneous velocity of the object at a general time t\ge 0. You should mimic the earlier example for the instantaneous velocity when s=-16t^2+100.

1. s=t^2

2. s=9.8t^2

3. s=-16t^2+2t

4. s=t^3

Ans.

1. 2t

2. 19.6t

3. -32t+2

4. 3t^2

Solution.

1.

The average velocity of the object over the interval [t,t+\Delta t] is \frac{\Delta s}{\Delta t}, so since s(t)=t^2:

\begin{aligned} \frac{\Delta s}{\Delta t} & = \frac{s(t+\Delta t)-s(t)}{\Delta t} \\ & = \frac{(t+\Delta t)^2 - t^2}{\Delta t} \\ & = \frac{(t^2+2t\Delta t+(\Delta t)^2)-(t^2)}{\Delta t} \\ & = \frac{2t\Delta t+(\Delta t)^2}{\Delta t} \\ & = \frac{\Delta t(2t+\Delta t)}{\Delta t} \\ & = 2t + \Delta t \end{aligned}

Now let the interval [t,t+\Delta t] get smaller and smaller indefinitely—that is let \Delta t get closer and closer to 0. Then the average velocity \frac{\Delta s}{\Delta t}=2t+\Delta t gets closer and closer to 2t+0=2t. Thus, the object has instantaneous velocity 2t at time t. This calculation can be interpreted as taking the limit of \frac{\Delta s}{\Delta t} as \Delta t approaches 0, written as follows:

\begin{aligned} & \qquad \textrm{instantaneous velocity at }t \\ & = \textrm{limit of average velocity over }[t,t+\Delta t]\textrm{ as }\Delta t\textrm{ approaches to }0 \\ & = \lim_{\Delta t\to 0}\frac{\Delta s}{\Delta t} \\ & = \lim_{\Delta t\to 0}(2t+\Delta t) \\ & = 2t+(0) \\ & = 2t \end{aligned}

2.

\begin{aligned} &\qquad \textrm{instantaneous velocity at }t\\ & = \lim_{\Delta t\to 0}\frac{\Delta s}{\Delta t}\\ & = \lim_{\Delta t\to 0}\frac{s(t+\Delta t)-s(t)}{\Delta t} \\ & = \lim_{\Delta t\to 0}\frac{9.8(t+\Delta t)^2 - 9.8t^2}{\Delta t} \\ & = \lim_{\Delta t\to 0}\frac{9.8(t^2+2t(\Delta t)+(\Delta t)^2) - 9.8t^2}{\Delta t} \\ & = \lim_{\Delta t\to 0}\frac{19.6t(\Delta t)+9.8(\Delta t)^2}{\Delta t} \\ & = \lim_{\Delta t\to 0}19.6t+9.8(\Delta t) \\ & = 19.6t+9.8(0) \\ & = 19.6t \end{aligned}

3.

\begin{aligned} &\qquad \textrm{instantaneous velocity at }t\\ & = \lim_{\Delta t\to 0}\frac{\Delta s}{\Delta t}\\ & = \lim_{\Delta t\to 0}\frac{s(t+\Delta t)-s(t)}{\Delta t} \\ & = \lim_{\Delta t\to 0}\frac{\big( -16(t+\Delta t)^2+2(t+\Delta t)\big) - (-16t^2+2t)}{\Delta t} \\ & = \lim_{\Delta t\to 0}\frac{-16(t^2+2t(\Delta t)+(\Delta t)^2)+2(t+\Delta t)+16t^2-2t}{\Delta t} \\ & = \lim_{\Delta t\to 0}\frac{-32t(\Delta t)-16(\Delta t)^2+2(\Delta t)}{\Delta t} \\ & = \lim_{\Delta t\to 0}(-32t-16(\Delta t)+2) \\ & = -32t-16(0)+2 \\ & = -32t+2 \end{aligned}

4.

\begin{aligned} s(t) & = t^3 \\ s(t+\Delta t) & = (t+\Delta t)^3 \\ & = t^3+3t^2(\Delta t)+3t(\Delta t)^2+(\Delta t)^3 \\ s(t+\Delta t)-s(t) & = \big( t^3+3t^2(\Delta t)+3t(\Delta t)^2+(\Delta t)^3 \big) - (t^3) \\ & = 3t^2(\Delta t)+3t(\Delta t)^2+(\Delta t)^3 \\ \frac{s(t+\Delta t)-s(t)}{\Delta t} & = \frac{3t^2(\Delta t)+3t(\Delta t)^2+(\Delta t)^3}{\Delta t} \\ & = 3t^2+3t(\Delta t)+(\Delta t)^2 \\ \lim_{\Delta t\to 0}\frac{s(t+\Delta t)-s(t)}{\Delta t} & = 3t^2+3t(0)+(0)^2 \\ \frac{\mathrm{d}s}{\mathrm{d}t} & = 3t^2\\ \end{aligned}