Find velocity, Calculus and rates of change

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Homework Help Overview

The problem involves finding the velocity of a particle based on its displacement given by the equation of motion s = 1/t², where t is in seconds. The original poster is tasked with determining the velocity at specific time points: t = a, t = 1, t = 2, and t = 3.

Discussion Character

  • Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to calculate velocity using the formula v = d/t but questions the negative values provided in the textbook's answers. Some participants suggest that the instantaneous velocity should be used instead of average velocity, prompting a discussion about derivatives.

Discussion Status

Participants are exploring the distinction between average and instantaneous velocity. Guidance has been offered regarding the need to apply the derivative to find instantaneous velocity, and there is ongoing clarification about the appropriate formulas to use.

Contextual Notes

There is a mention of the original poster's confusion regarding the negative values in the textbook's answers, indicating a potential misunderstanding of the problem setup or the application of calculus concepts.

physics604
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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.
 
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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) = ?
 

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