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vampirchik [111]
3 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]3 years ago
8 0
If it is s-t graph , point is c
if it is v-t graph , point is e
antiseptic1488 [7]3 years ago
7 0

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.

Therefore, the instantaneous velocity is zero at point C and G.

Learn more:

1. Describe velocity https://brainly.in/question/3480664

2. Average velocity brainly.com/question/190072

3. Difference between speed and velocity

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.

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schepotkina [342]

Answer:

thanks for the points

Explanation:

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laila [671]

Answer:

i) 21 cm

ii) At infinity behind the lens.

iii) A virtual, upright, enlarged image behind the object

Explanation:

First identify,

object distance (u) = 42 cm (distance between  object and lens, 50 cm - 8 cm)

image distance (v) = 42 cm (distance between  image and lens, 92 cm - 50 cm)

The lens formula,

\frac{1}{v} -\frac{1}{u} =\frac{1}{f}

Then applying the new Cartesian sign convention to it,

\frac{1}{v} +\frac{1}{u} =\frac{1}{f}

Where f is (-), u is (+) and  v is (-) in  all 3  cases. (If not values with signs have to considered, this method that need will not arise)

Substituting values you get,

i) \frac{1}{42} +\frac{1}{42} =\frac{1}{f}\\\frac{2}{42} =\frac{1}{f}

f = 21 cm

ii) u =21 cm, f = 21 cm v = ?

Substituting in same equation\frac{1}{v} =\frac{1}{21} =\frac{1}{21} \\\\\frac{1}{v} = 0\\

  v ⇒ ∞ and image will form behind the lens

iii) Now the object will be within the focal length of the lens. So like in the attachment, a virtual, upright, enlarged image behind the object.

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