A test 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

Question

vampirchik [111]

2 years ago

11

A test 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?

Physics

2 answers:

svetlana [45] · Answer 1 · 2022-08-01T02:13:43+03:00

svetlana [45]2 years ago

8 0

If it is s-t graph , point is c
if it is v-t graph , point is e

antiseptic1488 [7] · Answer 2 · 2022-08-01T03:44:40+03:00

The instantaneous velocity is zero at point $C$ and $G$ .

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, $v$ 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 $A$ :

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 $B$ :

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 $C$ :

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 $D$ :

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 $E$ :

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 $F$ :

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 $G$ :

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