Find velocity, Calculus and rates of change

In summary: 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}\
  • #1
physics604
92
2
1. The displacement (in meters) of a particle moving in a straight line is given by the equation of motion s=[itex]\frac{1}{t^2}[/itex], 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=[itex]\frac{d}{t}[/itex]

The Attempt at a Solution



The question says the d=[itex]\frac{1}{t^2}[/itex]

I plug that into V=[itex]\frac{d}{t}[/itex], getting s=[itex]\frac{1}{t^3}[/itex].

When I input t=a, 1, 2, and 3, I get [itex]\frac{1}{a^3}[/itex], 1, [itex]\frac{1}{8}[/itex], and [itex]\frac{1}{27}[/itex].

However, the textbook says that the answers are -[itex]\frac{2}{a^3}[/itex], -2, -[itex]\frac{1}{4}[/itex], and -[itex]\frac{2}{27}[/itex].

Where did it get the -2 from? What am I doing wrong?

Thanks in advance.
 
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  • #2
physics604 said:
1. The displacement (in meters) of a particle moving in a straight line is given by the equation of motion s=[itex]\frac{1}{t^2}[/itex], 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=[itex]\frac{d}{t}[/itex]

The Attempt at a Solution



The question says the d=[itex]\frac{1}{t^2}[/itex]

I plug that into V=[itex]\frac{d}{t}[/itex], getting s=[itex]\frac{1}{t^3}[/itex].

When I input t=a, 1, 2, and 3, I get [itex]\frac{1}{a^3}[/itex], 1, [itex]\frac{1}{8}[/itex], and [itex]\frac{1}{27}[/itex].

However, the textbook says that the answers are -[itex]\frac{2}{a^3}[/itex], -2, -[itex]\frac{1}{4}[/itex], and -[itex]\frac{2}{27}[/itex].

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.
 
  • #3
Thanks! According to my textbook, the equation for

instantaneous rates of change = lim x2→x1 [itex]\frac{f(x2)-f(x1)}{x2-x1}[/itex].

In my case would that mean the equation would be

[itex]\frac{Δd}{Δt}[/itex] = lim t2→t1 [itex]\frac{f(d2)-f(d1)}{t2-t1}[/itex] ?

How does that work into the question?
 
Last edited:
  • #4
$$v = ds/dt = \lim_{h \to 0}\frac{s(t + h) - s(t)}{h}$$

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) = ?
 

1. What is velocity?

Velocity is a measure of how fast an object is moving in a specific direction. It is a vector quantity, meaning it has both magnitude (speed) and direction.

2. How is velocity calculated using calculus?

Velocity can be calculated by taking the derivative of an object's position function with respect to time. This means finding the rate of change of position over time, also known as the slope of the position-time graph.

3. What is the difference between average velocity and instantaneous velocity?

Average velocity is the average rate of change of an object's position over a given time interval. Instantaneous velocity, on the other hand, is the velocity at a specific moment in time and is found by taking the limit of the average velocity as the time interval approaches zero.

4. How is calculus used to find rates of change?

Calculus is used to find rates of change by finding the derivative of a function. The derivative represents the rate of change of the function at a given point, and can be used to find instantaneous rates of change as well as average rates of change over a given interval.

5. Can calculus be used to find rates of change in real-world scenarios?

Yes, calculus is often used to find rates of change in real-world scenarios, such as calculating the speed of a moving object, the rate of growth of a population, or the rate of change of a stock price. It is a powerful tool for analyzing and understanding how quantities change over time.

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