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Instantaneous velocity is a fundamental concept in physics and is defined as the rate of change of an object’s position at a specific moment in time. Unlike average velocity, which considers the overall displacement over a given period, instantaneous velocity focuses on the precise velocity at a single point. This measurement can be vital in many fields, including physics, engineering, and even sports analysis. By understanding how to calculate instantaneous velocity, individuals can gain valuable insights into an object’s motion, including its direction, speed, and acceleration, at any given time. In this article, we will explore the steps and formulas involved in calculating instantaneous velocity, as well as the practical applications and significance of this concept in various contexts.
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Velocity is defined as the speed of an object in a specified direction. [1] X Research Source In many cases, to find the velocity we will use the equation v = s/t, where v is the velocity, s is the total distance traveled by the object from the initial position, and t is the time it takes the object to travel that distance. However, in theory this formula only gives the average velocity of the object over the distance. By calculus we can calculate the speed of the object at any time along the distance. That’s the instantaneous velocity and is defined by the equation v = (ds)/(dt) , or in other words, it’s the derivative of the equation for the average velocity. [2] X Research Source
Steps
Calculate instantaneous velocity
s = -1.5t 2 + 10t + 4
- In this equation, the variables are:
-
- s = displacement distance . Distance of the moving object from the starting position. For example, if an object travels 10 meters forward and 7 meters backward, its total distance traveled is 10 – 7 = 3 meters (not 10 + 7 = 17m).
- t = time . This variable simply needs no explanation, usually measured in seconds.
-
- In other words, start to differentiate from left to right on the “t” side of the equation. Every time you encounter the variable “t”, you subtract the exponent by 1 and multiply the whole term by the original exponent. Any constant terms (terms without “t”) will disappear because they are multiplied by 0. The process is actually not as difficult as you might think – let’s take the equation in the step above as an example:
s = -1.5t 2 + 10t + 4
(2)-1.5t (2-1) + (1)10t 1 – 1 + (0)4t 0
-3t 1 + 10t 0
-3t + 10
- In the example above, the derivative of the equation would look like this:
ds/dt = -3t + 10
ds/dt = -3t + 10
ds/dt = -3(5) + 10
ds/dt = -15 + 10 = -5 meters/second
- Note that we use the “meter/second” unit above. Since we are solving a problem with displacement in meters and time in seconds, where velocity is the distance traveled in time, this unit is appropriate.
Graph instantaneous velocity estimate
- To graph the distance traveled, you use the x-axis as the time and the y-axis as the distance traveled. You then determine a number of points by substituting the values of t into the equation of motion, the resulting s values, and you plot the points t,s (x,y) on the graph.
- Note that the graph can extend below the x-axis. If the line of motion of an object goes below the x-axis, this means that the object is moving backwards from its original position. In general, graphs won’t extend behind the y-axis – we don’t normally measure the velocity of an object moving backwards over time!
- Assume the distance traveled has points (1,3) and (4,7). In this case, if we want to find the slope at (1,3) we can set (1;3) = P and (4;7) = Q .
H = (y Q – y P )/(x Q – x P )
H = (7 – 3)/(4 – 1)
H = (4)/(3) = 1.33
Q = (2,4,8): H = (4.8 – 3)/(2 – 1)
H = (1,8)/(1) = 1.8Q = (1.5;3.95): H = (3.95 – 3)/(1.5 – 1)
H = (0.95)/(0.5) = 1.9Q = (1,25;3.49): H = (3.49 – 3)/(1.25 – 1)
H = (0.49)/(0.25) = 1.96
- In the above example, as we move H closer to P, we have H values of 1.8; 1.9 and 1.96. Since these numbers are approaching 2, we can say 2 is the approximate value of the slope at P.
- Remember that the slope at any point on the graph is the derivative of the equation of the graph at that point. Since the graph represents the distance traveled by an object over time, as we saw in the previous section, its instantaneous velocity at any given point is the derivative of the distance the object has moved at the point in question. approach, we can say that 2 meters/second is an approximate estimate of the instantaneous velocity when t = 1.
Sample problem
- First, take the derivative of the equation:
s = 5t 3 – 3t 2 + 2t + 9
s = (3)5t (3 – 1) – (2)3t (2 – 1) + (1)2t (1 – 1) + (0)9t 0 – 1
15t (2) – 6t (1) + 2t (0)
15t (2) – 6t + 2 - Then we substitute the value of t (4) in:
s = 15t (2) – 6t + 2
15(4) (2) – 6(4) + 2
15(16) – 6(4) + 2
240 – 24 + 2 = 22 meters/second
- First, we find the points Q when t = 2; 1.5; 1.1 and 1.01.
s = 4t 2 – t
t = 2: s = 4(2) 2 – (2)
4(4) – 2 = 16 – 2 = 14, so Q = (2,14)t = 1.5: s = 4(1.5) 2 – (1.5)
4(2.25) – 1.5 = 9 – 1.5 = 7.5, so Q = (1.5;7.5)t = 1,1: s = 4(1,1) 2 – (1,1)
4(1.21) – 1.1 = 4.84 – 1.1 = 3.74, so Q = (1,1;3.74)t = 1.01: s = 4(1.01) 2 – (1.01)
4(1.0201) – 1.01 = 4.0804 – 1.01 = 3.0704, so Q = (1.01;3.0704) - Next we will get the H values:
Q = (2,14): H = (14 – 3)/(2 – 1)
H = (11)/(1) = 11Q = (1.5;7.5): H = (7.5 – 3)/(1.5 – 1)
H = (4,5)/(0.5) = 9Q = (1,1;3.74): H = (3.74 – 3)/(1,1 – 1)
H = (0.74)/(0.1) = 7.3Q = (1.01;3.0704): H = (3.0704 – 3)/(1.01 – 1)
H = (0.0704)/(0.01) = 7.04 - Since the H values seem to approach 7, we can say that 7 m/s is an approximate estimate of the instantaneous velocity at coordinates (1;3).
Advice
- To find the acceleration (change in velocity with respect to time), use the method in part one to take the derivative of the displacement equation. Then take the derivative again for the derivative equation just found. As a result, you have an equation that finds the acceleration at a given time – all you have to do is plug in the time value.
- The equation representing the correlation between Y (displacement) and X (time) can be very simple, like Y = 6x + 3. In this case, the slope is constant and it is not necessary to take it. derivative to calculate the slope, that is, it follows the form of the basic equation Y = mx + b for a graph of a linear line, ie the slope is 6.
- Distance traveled is like distance but has direction, so it is a vector quantity, and speed is a scalar quantity. Distance traveled can be negative, while distance is only positive.
wikiHow is a “wiki” site, which means that many of the articles here are written by multiple authors. To create this article, 25 people, some of whom are anonymous, have edited and improved the article over time.
This article has been viewed 92,696 times.
Velocity is defined as the speed of an object in a specified direction. [1] X Research Source In many cases, to find the velocity we will use the equation v = s/t, where v is the velocity, s is the total distance traveled by the object from the initial position, and t is the time it takes the object to travel that distance. However, in theory this formula only gives the average velocity of the object over the distance. By calculus we can calculate the speed of the object at any time along the distance. That’s the instantaneous velocity and is defined by the equation v = (ds)/(dt) , or in other words, it’s the derivative of the equation for the average velocity. [2] X Research Source
In conclusion, calculating instantaneous velocity is a crucial step in understanding the motion of an object at a specific moment in time. By using the concept of limits and taking the derivative of the position function with respect to time, we can determine the rate at which an object is moving at a particular instant. This calculation involves finding the slope of a tangent line to the position-time graph, which represents the object’s velocity at that exact point. Through the use of calculus and various mathematical techniques, we can accurately determine the instantaneous velocity of an object, providing valuable insights into its motion and behavior. Understanding this concept is essential for scientists, engineers, and anyone interested in studying or analyzing motion in a precise and detailed manner.
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