# Find velocity, Calculus and rates of change

• physics604
Therefore,$$v = \lim_{h \to 0}\frac{s(t + h) - s(t)}{h} = \lim_{h \to 0}\frac{\frac{1}{(t + h)^2} - \frac{1}{t^2}}{h} = \lim_{h \to 0}\frac{\frac{t^2 - (t + h)^2}{(t^2)(t + h)^2}}{h} = \lim_{h \to 0}\ #### physics604 1. The displacement (in meters) of a particle moving in a straight line is given by the equation of motion s=$\frac{1}{t^2}$, where t is measured in seconds. Find the velocity of the particle at times t=a, t=1, t=2, and t=3. ## Homework Equations v=$\frac{d}{t}$ ## The Attempt at a Solution The question says the d=$\frac{1}{t^2}$ I plug that into V=$\frac{d}{t}$, getting s=$\frac{1}{t^3}$. When I input t=a, 1, 2, and 3, I get $\frac{1}{a^3}$, 1, $\frac{1}{8}$, and $\frac{1}{27}$. However, the textbook says that the answers are -$\frac{2}{a^3}$, -2, -$\frac{1}{4}$, and -$\frac{2}{27}$. Where did it get the -2 from? What am I doing wrong? Thanks in advance. physics604 said: 1. The displacement (in meters) of a particle moving in a straight line is given by the equation of motion s=$\frac{1}{t^2}$, where t is measured in seconds. Find the velocity of the particle at times t=a, t=1, t=2, and t=3. ## Homework Equations v=$\frac{d}{t}$ ## The Attempt at a Solution The question says the d=$\frac{1}{t^2}$ I plug that into V=$\frac{d}{t}$, getting s=$\frac{1}{t^3}$. When I input t=a, 1, 2, and 3, I get $\frac{1}{a^3}$, 1, $\frac{1}{8}$, and $\frac{1}{27}$. However, the textbook says that the answers are -$\frac{2}{a^3}$, -2, -$\frac{1}{4}$, and -$\frac{2}{27}$. Where did it get the -2 from? What am I doing wrong? The formula you are using is the average velocity. In the problem, you're supposed to use the instantaneous velocity, ##\frac{ds}{dt}##. Your book should have examples of how to find the derivative of the displacement, s. Thanks! According to my textbook, the equation for instantaneous rates of change = lim x2→x1 $\frac{f(x2)-f(x1)}{x2-x1}$. In my case would that mean the equation would be $\frac{Δd}{Δt}$ = lim t2→t1 $\frac{f(d2)-f(d1)}{t2-t1}$ ? How does that work into the question? Last edited:$$v = ds/dt = \lim_{h \to 0}\frac{s(t + h) - s(t)}{h}

This formula is equivalent to the one in your book, with x2 = t + h and x1 = t.

Your function is s = s(t) = 1/t2.

So s(t + h) = ?
And s(t) = ?