The instantaneous velocity is zero at point [tex]C[/tex] and [tex]G[/tex].
Further Explanation:
The velocity of an object in motion at a specific time and at specific point is termed as instantaneous velocity. Along a path of motion the given points can be separated by some finite distance.
Concept:
The expression for the instantaneous velocity is:
[tex]\fbox{\begin\\v=\dfrac{{{x_1} - {x_o}}}{{{t_1} - {t_o}}}\end{minispace}}[/tex]
Here, [tex]v[/tex] is the velocity, [tex]{x_o}[/tex] is the initial position, [tex]{x_1}[/tex] is the final position, [tex]{t_1}[/tex] is the final time and [tex]{t_o}[/tex] is the initial time.
For instantaneous velocity at point [tex]A[/tex]:
Substitute [tex]20\text{ m}[/tex] for [tex]{x_o}[/tex], [tex]40\text{ m}[/tex] for [tex]{x_1}[/tex], [tex]0\text{ s}[/tex] for [tex]{t_o}[/tex] and [tex]3\text{ s}[/tex] for [tex]{t_1}[/tex] in the above equation.
[tex]\begin{aligned}\\v&=\frac{{40{\text{ m}}-20{\text{ m}}}}{{3{\text{ s}}-0{\text{ s}}}}\\&=\frac{{20}}{3}{\text{ m/s}}\\&=6.67{\text{ m/s}}\\\end{aligned}[/tex]
For instantaneous velocity at point [tex]B[/tex]:
Substitute [tex]20\text{ m}[/tex] for [tex]{x_o}[/tex], [tex]40\text{ m}[/tex] for [tex]{x_1}[/tex], [tex]0\text{ s}[/tex] for [tex]{t_o}[/tex] and [tex]3\text{ s}[/tex] for [tex]{t_1}[/tex] in the above equation.
[tex]\begin{aligned}\\v&=\frac{{40{\text{ m}}-20{\text{ m}}}}{{3{\text{ s}}-0{\text{ s}}}}\\&=\frac{{20}}{3}{\text{ m/s}}\\&=6.67{\text{ m/s}}\\\end{aligned}[/tex]
For instantaneous velocity at point [tex]C[/tex]:
Substitute [tex]40\text{ m}[/tex] for [tex]{x_o}[/tex], [tex]40\text{ m}[/tex] for [tex]{x_1}[/tex], [tex]3\text{ s}[/tex] for [tex]{t_o}[/tex] and [tex]5\text{ s}[/tex] for [tex]{t_1}[/tex] in the above equation.
[tex]\begin{aligned}v&=\frac{{40{\text{ m}}-40{\text{ m}}}}{{{\text{5 s}}-3{\text{ s}}}}\\&=\frac{{-0{\text{ m}}}}{2}{\text{ m/s}}\\&=0{\text{ m/s}}\\ \end{aligned}[/tex]
For instantaneous velocity at point [tex]D[/tex]:
Substitute [tex]40\text{ m}[/tex] for [tex]{x_o}[/tex], [tex]0\text{ m}[/tex] for [tex]{x_1}[/tex], [tex]5\text{ s}[/tex] for [tex]{t_o}[/tex] and [tex]6\text{ s}[/tex] for [tex]{t_1}[/tex] in the above equation.
[tex]\begin{aligned}v&=\frac{{0{\text{ m}}-40{\text{ m}}}}{{{\text{6 s}}-5{\text{ s}}}}\\&=\frac{{-40{\text{ m}}}}{1}{\text{ m/s}} \\ &=-40{\text{ m/s}}\\ \end{aligned}[/tex]
For instantaneous velocity at point [tex]E[/tex]:
Substitute [tex]40\text{ m}[/tex] for [tex]{x_o}[/tex], [tex]0\text{ m}[/tex] for [tex]{x_1}[/tex], [tex]5\text{ s}[/tex] for [tex]{t_o}[/tex] and [tex]6\text{ s}[/tex] for [tex]{t_1}[/tex] in the above equation.
[tex]\begin{aligned}v&=\frac{{0{\text{ m}}-40{\text{ m}}}}{{{\text{6 s}}-5{\text{ s}}}}\\&=\frac{{-40{\text{ m}}}}{1}{\text{ m/s}}\\&=-40{\text{ m/s}}\\ \end{aligned}[/tex]
For instantaneous velocity at point [tex]F[/tex]:
Substitute [tex]40\text{ m}[/tex] for [tex]{x_o}[/tex], [tex]0\text{ m}[/tex] for [tex]{x_1}[/tex], [tex]5\text{ s}[/tex] for [tex]{t_o}[/tex] and [tex]6\text{ s}[/tex] for [tex]{t_1}[/tex] in the above equation.
[tex]\begin{aligned}v&=\frac{{0{\text{ m}}-40{\text{ m}}}}{{{\text{6 s}}-5{\text{ s}}}}\\&=\frac{{-40{\text{ m}}}}{1}{\text{ m/s}}\\&=-40{\text{ m/s}}\\ \end{aligned}[/tex]
For instantaneous velocity at point [tex]G[/tex]:
The instantaneous velocity is [tex]0\text{ m/s}[/tex], because slope at on the curve at point [tex]G[/tex] is parallel to the time axis.
Therefore, the instantaneous velocity is zero at point [tex]C[/tex] and [tex]G[/tex].
Learn more:
1. Describe velocity https://brainly.in/question/3480664
2. Average velocity https://brainly.com/question/190072
3. Difference between speed and velocity
https://brainly.com/question/2880714
Answer Details:
Grade: Middle school
Subject: Physics
Chapter: Kinematics
Keywords:
Test, car travels, straight, line, x-axis, graph, figure, shows, position, function, time, velocity, zero, point C and G.
