Find velocity, Calculus and rates of change

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SUMMARY

The discussion focuses on finding the velocity of a particle described by the displacement equation s = 1/t². The user initially applies the average velocity formula v = d/t incorrectly, leading to incorrect results for specific times t = a, 1, 2, and 3. The correct approach involves using the instantaneous velocity formula v = ds/dt, which requires calculating the derivative of the displacement function. The textbook provides the correct answers as -2/a³, -2, -1/4, and -2/27, highlighting the importance of understanding derivatives in calculus.

PREREQUISITES
  • Understanding of calculus concepts, particularly derivatives.
  • Familiarity with the displacement equation s = 1/t².
  • Knowledge of instantaneous vs. average velocity.
  • Ability to apply limits in calculus.
NEXT STEPS
  • Learn how to differentiate functions using the power rule.
  • Study the concept of instantaneous velocity and its applications.
  • Explore the limit definition of derivatives in calculus.
  • Practice solving problems involving rates of change and displacement equations.
USEFUL FOR

Students studying calculus, particularly those focusing on motion and rates of change, as well as educators seeking to clarify concepts of instantaneous velocity and derivatives.

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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