Instant speed: definition, formula, calculation and exercises

To get the instantaneous speed, let’s look at a time span as short as possible. Differential calculus is the perfect tool to express this idea mathematically.

The starting point is the average speed:

This limit is known as a derivative. In differential calculus notation, you have:

Whenever motion is restricted to a straight line, vector notation can be dispensed with.

Instant velocity calculation: geometric interpretation

The following figure shows the geometric interpretation of the derivative concept: it is the slope of the tangent line to the curve x (t) vs. t at each point.

The instantaneous velocity at P is numerically equivalent to the slope of the tangent line to the x vs curve. t at point P. Source: Source: CC0 じ に く シ チ ュ [CC0].

You can figure out how to get the limit if you gradually approach point Q to point P. There will come a time when the two points will be so close that you won’t be able to distinguish one from the other.

The line joining them will change from secant (line that cuts at two points) to tangent (line that touches the curve at a single point). Therefore, to find the instantaneous velocity of a moving particle, we should have:

• The plot of particle position versus time. By finding the slope of the line tangent to the curve at each instant of time, you get the instantaneous velocity at each point occupied by the particle.

The good:

• The particle position function x(t) , which is derived to obtain the velocity function v(t) , then this function is evaluated every time t , for convenience. The position function is supposed to be derivable.

Some special cases in instantaneous velocity calculation

-The slope of the line tangent to the curve at P is 0. A null slope means that the rover is stationary and its velocity is certainly 0.

-The slope of the line tangent to the curve at P is greater than 0. Velocity is positive. In the graph above, this means that the cell is moving away from O.

-The slope of the line tangent to the curve at P is less than 0. The velocity would be negative. In the graph above, there are no such points, but in this case the particle would be approaching O.

-The slope of the line tangent to the curve is constant at P and all other points. In this case, the graph is a straight line and the cell phone has MRU of uniform rectilinear movement (its velocity is constant).

In general, the function v(t) is also a function of time, which in turn can have a derivative. What if it was not possible to find the derivatives of the functions x (t) and v (t) ?

In the case of x (t), it may be that the slope – instantaneous velocity – changes sign abruptly. Or that it will go from zero to a different value immediately.

In this case, the x(t) graph would have points or corners at the locations of sudden changes. Very different from the case represented in the previous image, where the x (t) curve is a smooth curve, with no points, corners, discontinuities or sudden changes.

The truth is that, for real cell phones, smooth curves are the ones that best represent the object’s behavior.

The movement in general is quite complex. Cell phones can be stopped for a while, accelerate to go from rest to get a speed away from the starting point, hold the speed for a while and then brake to stop again, and so on.

Again they can start again and continue in the same direction. Or activate indentation and return. This is called motion varied in one dimension.

Here are some examples of instantaneous velocity calculation to clarify the use of the provided definitions:

Solved Instant Speed ​​Exercises

Exercise 1

A particle travels along a straight line with the following law of motion:

x (t) = -t ³ + 2 t ² + 6 t – 10

All units are in the international system. To locate:

a) The position of the particle at t = 3 seconds.

b) The average velocity in the interval between t = 0 and t = 3 s.

c) The mean velocity in the interval between t = 0 and t = 3 s.

d) The instantaneous velocity of the particle from the previous question, at t = 1 s.

Answers

a) To find the particle’s position, the law of motion (position function) is evaluated at t = 3:

x (3) = (-4/3) .3 ³ + 2. 3 ² ₊ 6.3 – 10 m = -10 m

It is okay for the position to be negative. The sign (-) indicates that the particle is to the left of the O origin.

b) In calculating the average velocity, the final and initial positions of the particle at the indicated times are required: x (3) and x (0). The position at t = 3 is x (3) and the previous result is known. The position at t = 0 seconds is x (0) = -10 m.

Since the end position is equal to the start position, it is immediately concluded that the average velocity is 0.

c) Average speed is the ratio between the distance covered and the time spent. Now, distance is the magnitude or magnitude of the displacement, so:

distance = | x2 – x1 | = | -10 – (-10) | m = 20 m

Note that the distance covered is always positive.

v _{m = 20 m / 3 s = 6.7 m / s}

d) Here it is necessary to find the first derivative of position with respect to time. It is then evaluated by t = 1 second.

x ‘(t) = -4 t ² + 4 t + 6

x ‘(1) = -4.1 ² + 4.1 + 6 m / s = 6 m / s

Exercise 2

Below is a graph of the position of a cell phone as a function of time. Find the instantaneous velocity at t = 2 seconds.

Response

Draw the line tangent to the curve at t = 2 seconds and calculate its slope by taking two points on the line.

In this example, we will take two points that are easily visualized, whose coordinates are (2 s, 10 m) and the cut with the vertical axis (0 s, 7 m):