Respuesta :
Velocity is defined as the change in position per unit time, which is literally the slope of the position vs time graph. Thus, the instantaneous velocity at point B is the slope of the graph at point B, which is approximately
(40m - 20m) / (3s - 0s)
= 6.67 m/s.
The instantaneous velocity of the car at position B is [tex]\boxed{6.67{\text{ m/s}}}[/tex].
Further Explanation:
The instantaneous velocity for a position-time straight line graph is defined as the mathematical slope of the graph. Or specifically for a segment of the graph, it is the slope of that segment.
The slope remains same for all the points on the continuous segment of the graph. So, here in the given question, whatever be the slope (velocity), it will remain same for every point on a particular segment. For example: in segment DE, every point on the segment DE has the same value of velocity including points D and E. Or for segment AB same thing follows. So, whatever be the slope (velocity) of the line will be the same for points A and B.
Mathematically, the slope is defined as the tangent (trigonometry) of the angle made by the considered segment with the x axis or it is also defined as the ratio of change in the quantity on y axis to change in the quantity on x axis.
[tex]slope = \tan \left( \text{angle made by the considered segment of graph with x axis} \right)[/tex]
[tex]slope=\dfrac{\text{change in quantity given on y axis}}{\text{change in quantity given on x axis}}[/tex]
Concept:
The slope of position-time graph gives velocity.
Therefore, by the definition of slope, we have
[tex]velocity=\dfrac{{change\,in\,position}}{{change\,in\,time}}[/tex]
Now taking two points on the graph for finding the slope (velocity) by point slope form which is as follows:
[tex]slope=\dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}[/tex]
So, the velocity as slope will be given as:
[tex]velocity = \dfrac{{{p_2} -{p_1}}}{{{t_2} - {t_1}}}[/tex]
Substituting [tex]40\,{\text{m}}[/tex] for [tex]p_2[/tex], [tex]20\,{\text{m}}[/tex] for [tex]p_1}[/tex], [tex]3\,{\text{s}}[/tex] for [tex]{t_2}[/tex], [tex]0\,{\text{s}}[/tex] for [tex]{t_1}[/tex].
[tex]\begin{aligned}v&=\frac{{40{\text{ m -20 m}}}}{{3{\text{ s -0 s}}}}\\&=\frac{{20{\text{ m}}}}{{3{\text{ s}}}}\\&=6.67{\text{ m/s}}\\\end{aligned}[/tex]
Therefore, the instantaneous velocity of the car at point B is [tex]\boxed{6.67{\text{ m/s}}}[/tex].
Learn More:
1. Conservation of momentum https://brainly.com/question/9484203
2. What fraction of the rope can hang over the edge of the table without the rope sliding https://brainly.com/question/2959748
Answer Details:
Grade: Middle school
Subject: Physics
Chapter: kinematics
Keywords:
Instantaneous velocity, car, velocity, test car, straight line, x axis, 40m, 20 m, 3 s, 0 s, position, time, position-time graph, at point B, 6.67 m/s, 6.7 m/s.
