Summarize this article in one sentence.
To get an object's instantaneous velocity, first we have to have an equation that tells us its position (in terms of displacement) at a certain point in time. This means the equation must have the variable s on one side by itself and t on the other (but not necessarily by itself), like this:s = -1.5t2 + 10t + 4 In this equation, the variables are:   Displacement = s . The distance the object has traveled from its starting position. For example, if an object goes 10 meters forward and 7 meters backward, its total displacement is 10 - 7 = 3 meters (not 10 + 7 = 17 meters).  Time = t . Self explanatory. Typically measured in seconds. The derivative of an equation is just a different equation that tells you its slope at any given point in time. To find the derivative of your displacement formula, differentiate the function with this general rule for finding derivatives: If y = a*xn, Derivative = a*n*xn-1.This rule is applied to every term on the "t" side of the equation. In other words, start by going through the "t" side of your equation from left to right. Every time you reach a "t", subtract 1 from the exponent and multiply the entire term by the original exponent. Any constant terms (terms which don't contain "t") will disappear because they be multiplied by 0. This process isn't actually as hard as it sounds — let's derive the equation in the step above as an example:s = -1.5t2 + 10t + 4(2)-1.5t(2-1) + (1)10t1 - 1 + (0)4t0-3t1 + 10t0-3t + 10 " To show that our new equation is a derivative of the first one, we replace "s" with the notation "ds/dt". Technically, this notation means "the derivative of s with respect to t." A simpler way to think of this is just that ds/dt is just the slope of any given point in the first equation. For example, to find the slope of the line made by s = -1.5t2 + 10t + 4 at t = 5, we would just plug "5" into t in its derivative. In our running example, our finished equation should now look like this:ds/dt = -3t + 10 Now that you have your derivative equation, finding the instantaneous velocity at any point in time is easy. All you need to do is pick a value for t and plug it into your derivative equation. For example, if we want to find the instantaneous velocity at t = 5, we would just substitute "5" for t in the derivative ds/dt = -3 + 10. Then, we'd just solve the equation like this:ds/dt = -3t + 10ds/dt = -3(5) + 10ds/dt = -15 + 10 = -5 meters/second Note that we use the label "meters/second" above. Since we're dealing with displacement in terms of meters and time in terms of seconds and velocity in general is just displacement over time, this label is appropriate.
Start with an equation for velocity in terms of displacement. Take the equation's derivative. Replace "s" with "ds/dt. Plug in a t value for your new equation to find instantaneous velocity.