Physics, 13.07.2019 08:00 anyone2562

Atest car travels in a straight line along the x-axis. the graph in the figure shows the car’s position x as a function of time. at which point is the velocity zero?

Atest car travels in a straight line along the x-axis. the graph in the figure shows the car’s posit

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Answer from: elizabethseoane6321

The instantaneous velocity is zero at point and .

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:

$\fbox{\begin\\v=\dfrac{{{x_1} - {x_o}}}{{{t_1} - {t_o}}}\end{minispace}}$

Here, is the velocity, ${x_o}$ is the initial position, ${x_1}$ is the final position, ${t_1}$ is the final time and ${t_o}$ is the initial time.

For instantaneous velocity at point :

Substitute $20\text{ m}$ for ${x_o}$ , $40\text{ m}$ for ${x_1}$ , $0\text{ s}$ for ${t_o}$ and $3\text{ s}$ for ${t_1}$ in the above equation.

$\begin{aligned}\\v&=\frac{{40{\text{ m}}-20{\text{ m}}}}{{3{\text{ s}}-0{\text{ s}}}}\\&=\frac{{20}}{3}{\text{ m/s}}\\&=6.67{\text{ m/s}}\\\end{aligned}$

For instantaneous velocity at point :

Substitute $20\text{ m}$ for ${x_o}$ , $40\text{ m}$ for ${x_1}$ , $0\text{ s}$ for ${t_o}$ and $3\text{ s}$ for ${t_1}$ in the above equation.

$\begin{aligned}\\v&=\frac{{40{\text{ m}}-20{\text{ m}}}}{{3{\text{ s}}-0{\text{ s}}}}\\&=\frac{{20}}{3}{\text{ m/s}}\\&=6.67{\text{ m/s}}\\\end{aligned}$

For instantaneous velocity at point :

Substitute $40\text{ m}$ for ${x_o}$ , $40\text{ m}$ for ${x_1}$ , $3\text{ s}$ for ${t_o}$ and $5\text{ s}$ for ${t_1}$ in the above equation.

$\begin{aligned}v&=\frac{{40{\text{ m}}-40{\text{ m}}}}{{{\text{5 s}}-3{\text{ s}}}}\\&=\frac{{-0{\text{ m}}}}{2}{\text{ m/s}}\\&=0{\text{ m/s}}\\ \end{aligned}$

For instantaneous velocity at point :

Substitute $40\text{ m}$ for ${x_o}$ , $0\text{ m}$ for ${x_1}$ , $5\text{ s}$ for ${t_o}$ and $6\text{ s}$ for ${t_1}$ in the above equation.

$\begin{aligned}v&=\frac{{0{\text{ m}}-40{\text{ m}}}}{{{\text{6 s}}-5{\text{ s}}}}\\&=\frac{{-40{\text{ m}}}}{1}{\text{ m/s}} \\ &=-40{\text{ m/s}}\\ \end{aligned}$

For instantaneous velocity at point :

Substitute $40\text{ m}$ for ${x_o}$ , $0\text{ m}$ for ${x_1}$ , $5\text{ s}$ for ${t_o}$ and $6\text{ s}$ for ${t_1}$ in the above equation.

$\begin{aligned}v&=\frac{{0{\text{ m}}-40{\text{ m}}}}{{{\text{6 s}}-5{\text{ s}}}}\\&=\frac{{-40{\text{ m}}}}{1}{\text{ m/s}}\\&=-40{\text{ m/s}}\\ \end{aligned}$

For instantaneous velocity at point :

Substitute $40\text{ m}$ for ${x_o}$ , $0\text{ m}$ for ${x_1}$ , $5\text{ s}$ for ${t_o}$ and $6\text{ s}$ for ${t_1}$ in the above equation.

$\begin{aligned}v&=\frac{{0{\text{ m}}-40{\text{ m}}}}{{{\text{6 s}}-5{\text{ s}}}}\\&=\frac{{-40{\text{ m}}}}{1}{\text{ m/s}}\\&=-40{\text{ m/s}}\\ \end{aligned}$

For instantaneous velocity at point :

The instantaneous velocity is $0\text{ m/s}$ , because slope at on the curve at point is parallel to the time axis.

Therefore, the instantaneous velocity is zero at point and .

Learn more:

1. Describe velocity

2. Average velocity

3. Difference between speed and velocity

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.

Answer from: Quest

the awnser is d you right bro

Answer from: Quest

Since you have not included the experiment situation and the data, of course, i can not give you specific details about it. but i can you with this question.1) i will assume that the motion is projectile (parabolic) motion.2) yes, the horizontal distance depends on the horizontal velocity as, naming vx the horizontal velocity and t the time, it is: x = vx × t3) now, observe that the flight time depends on the vertical velocity as, naming voy the initial vertical velocity, t the time and y the vertical distance before reaching the ground: y = yo + voy×t - gt² / 2.the higher the initial vertical velocity (voy) the longer the object will be flying and so t will be higher in the equation for the horizontal distance.that explains the relationship between the horizontal distance and the horizontal velocity.

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