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How to Calculate Instantaneous Velocity

September 27, 2023 by admin Category: How To

You are viewing the article How to Calculate Instantaneous Velocity  at Tnhelearning.edu.vn you can quickly access the necessary information in the table of contents of the article below.

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

Table of Contents

  • Steps
    • Calculate instantaneous velocity
    • Graph instantaneous velocity estimate
    • Sample problem
  • Advice

Steps

Calculate instantaneous velocity

Image titled Calculate Instantaneous Velocity Step 1

Image titled Calculate Instantaneous Velocity Step 1

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Start with the equation for velocity versus distance traveled. To find the instantaneous velocity, we must first have an equation that shows the position of the object (in terms of distance traveled) at any given time. That is, the equation must have only one variable s on one side and a variable t on the other side (not necessarily only one variable), like this:

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.
Image titled Calculate Instantaneous Velocity Step 2

Image titled Calculate Instantaneous Velocity Step 2

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Take the derivative of the equation. The derivative of the equation is another equation that tells the slope of the distance at a particular time. To find the derivative of the equation with respect to the displacement distance, differentiate the function according to the following general rule to calculate the derivative: If y = a*x n , Derivative = a*n*x n-1 . This principle applies to every term on the “t” side of the equation.

  • 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

Image titled Calculate Instantaneous Velocity Step 3

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Image titled Calculate Instantaneous Velocity Step 3

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Replace “s” with “ds/dt”. To represent the new equation as the derivative of the original square, we replace “s” with the symbol “ds/dt”. In theory, this notation is “the derivative of s with respect to t”. A simpler way to understand this notation, ds/dt is the slope of any point in the original equation. For example, to find the slope of the distance described by the equation s = -1.5t 2 + 10t + 4 at time t = 5, we substitute “5” for t in the derivative of the equation.

  • In the example above, the derivative of the equation would look like this:

    ds/dt = -3t + 10

Image titled Calculate Instantaneous Velocity Step 4

Image titled Calculate Instantaneous Velocity Step 4

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Substitute a value of t into the new equation to find the instantaneous velocity. Now that we have the derivative equation, finding the instantaneous velocity at any instant is very easy. All you need to do is choose a value of t and substitute in the derivative equation. For example, if we want to find the instantaneous velocity at t = 5, we just need to substitute “5” for t in the derivative equation ds/dt = -3t + 10. We will solve the equation as follows:

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

Image titled Calculate Instantaneous Velocity Step 5

Image titled Calculate Instantaneous Velocity Step 5

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Plot the distance of the object against time. In the previous section, we said that the derivative is also a formula that allows us to find the slope at any point of the equation to be derived. In fact, if you plot the distance the object has moved on a graph, the slope of the graph at any given point is the instantaneous velocity of the object at that point .

  • 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!
Image titled Calculate Instantaneous Velocity Step 6

Image titled Calculate Instantaneous Velocity Step 6

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Choose a point P and a point Q located near point P on the graph. To find the slope of the graph at point P, we use the technique of “finding the limit”. Finding the limit means taking two points (P and Q (a point located near P)) on the curve and finding the slope of the line connecting those two points, repeating this process as the distance between P and Q shortens gradually.

  • 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 .
Image titled Calculate Instantaneous Velocity Step 7

Image titled Calculate Instantaneous Velocity Step 7

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Find the slope between P and Q. The slope between P and Q is the difference of the y values for P and Q over the difference of the x values for P and Q. In other words, H = (y Q – y P )/(x Q – x P ) , where H is the slope between two points. In this example, the slope between P and Q is:

H = (y Q – y P )/(x Q – x P )
H = (7 – 3)/(4 – 1)
H = (4)/(3) = 1.33

Image titled Calculate Instantaneous Velocity Step 8

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Image titled Calculate Instantaneous Velocity Step 8

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Repeat several times by moving Q closer to P. The goal is to reduce the distance between P and Q until they come close to a single point. The smaller the distance between P and Q, the closer the slope of that infinitesimal line will be to the slope at point P. Repeat a few times for our example equation, using points (2;4 ,8), (1.5,3.95) and (1.25;3.49) for Q and the initial coordinates of P are (1,3):

Q = (2,4,8): H = (4.8 – 3)/(2 – 1)
H = (1,8)/(1) = 1.8

Q = (1.5;3.95): H = (3.95 – 3)/(1.5 – 1)
H = (0.95)/(0.5) = 1.9

Q = (1,25;3.49): H = (3.49 – 3)/(1.25 – 1)
H = (0.49)/(0.25) = 1.96

Image titled Calculate Instantaneous Velocity Step 9

Image titled Calculate Instantaneous Velocity Step 9

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Estimate the slope of the infinitesimal line segment on the graph curve. As Q gets closer and closer to P, H will gradually get closer and closer to the slope at P. Finally, at an infinitely small line segment, H will be the slope at P. Since we cannot measure or calculate the length of a line segment is extremely small, so the slope at P is only estimated when that value is obvious from the points we compute.

  • 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

Image titled Calculate Instantaneous Velocity Step 10

Image titled Calculate Instantaneous Velocity Step 10

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Find the instantaneous velocity at t = 1 with the displacement equation s = 5t 3 – 3t 2 + 2t + 9. Same as the example in the first part but this is a cubic equation instead of a 2nd order, so we We can solve the problem in a similar way.

  • 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

  • Image titled Calculate Instantaneous Velocity Step 11

    Image titled Calculate Instantaneous Velocity Step 11

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    Use graphical estimation to find the instantaneous velocity at (1,3) for the displacement equation s = 4t 2 – t. For this problem, we use the coordinate (1,3) as the point P, but we have to find other Q points located near it. Then all we need to do is find the H values and deduce the estimate.

    • 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) = 11

      Q = (1.5;7.5): H = (7.5 – 3)/(1.5 – 1)
      H = (4,5)/(0.5) = 9

      Q = (1,1;3.74): H = (3.74 – 3)/(1,1 – 1)
      H = (0.74)/(0.1) = 7.3

      Q = (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).
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  • 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.
    X

    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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