The displacement (in meters) of a particle moving in a...

The displacement (in meters) of a particle moving in a straight line is given by the equation of motion
$ s = 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 $.

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Key Concepts

Velocity as the Derivative of Displacement

In kinematics, velocity is defined as the derivative of the displacement with respect to time. By differentiating the displacement function, one obtains the velocity function, which allows for evaluation of the object's speed and direction at any given moment.

Power Rule

The power rule is a fundamental rule in differentiation used for functions of the form t^n. It states that the derivative of t^n is n*t^(n-1). This rule simplifies the process of differentiating polynomial expressions, including those that involve negative exponents.

Differentiation

Differentiation is the process of finding the derivative of a function, which measures how the function changes in response to changes in its input. In physics, it is used to determine the instantaneous rate of change, such as the velocity of an object from its displacement function.

Evaluating Functions at Specific Points

After obtaining a derivative, evaluating it at particular values of the variable gives the rate of change at those points. In the context of the problem, this means calculating the velocity at predetermined times by substituting those time values into the derivative function.

Transcript

00:01 Suppose the position of a particle at time t is defined by the function s of t, which is equal to one over t squared.

00:08 And here we want to find the velocity at time t, which is equal to a 1, 2 and three to do this, we find the derivative of s at a point t, which is equal to a.

00:23 Now by definition of the derivative at the point we have s prime of a.

00:28 This is equal to limit as t approaches a.

00:34 Of s f t minus s.

00:36 Of a fish all over t minus a.

00:41 So from here we have limits.

00:43 S t approaches a.

00:44 Of s f t, which is one over t squared.

00:47 This minus s of a.

00:49 Which is one over a squared all over t minus a.

00:57 Now combining the numerator, we have limit s.

01:01 T approaches a.

01:03 Of we have a common denominator of a square t squared and then we have a squared minus d squared.