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Calculating instantaneous velocity is a fundamental concept in physics that helps us understand an object’s motion in a precise manner. While average velocity provides a general understanding of an object’s speed over a given interval, instantaneous velocity provides information about an object’s exact speed at a specific point in time. To accurately calculate instantaneous velocity, one must take into account the concept of limits and derivatives. By examining an object’s position over incredibly small time intervals, we can determine its instantaneous velocity at any given moment. In this article, we will explore the steps involved in calculating instantaneous velocity, as well as the significance of this concept in understanding the motion of objects in the physical world.
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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 requires understanding the basic principles of calculus and the concept of limits. By taking the derivative of a function describing the position of an object with respect to time, we can find the derivative at a specific instant, representing the instantaneous velocity at that moment. This calculation allows for a more precise understanding of an object’s motion, capturing changes in velocity over small time intervals. Instantaneous velocity is a crucial concept in physics and engineering, providing valuable insights into the behavior of objects in motion. Mastering this calculation method is essential in analyzing and predicting the movement of objects in various scenarios, from simple one-dimensional motion to complex systems.
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