(A) The graph is a circle with center $(2, 0)$ and radius $2 \ m$. Its equation is given by:$(s-2)^2 + v^2 = 2^2$$(s-2)^2 + v^2 = 4$Differentiating bo

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(A) The graph is a circle with center $(2, 0)$ and radius $2 \ m$. Its equation is given by:$(s-2)^2 + v^2 = 2^2$$(s-2)^2 + v^2 = 4$Differentiating both sides with respect to time $t$,we get:$2(s-2) \frac{ds}{dt} + 2v \frac{dv}{dt} = 0$Since $\frac{ds}{dt} = v$ and $\frac{dv}{dt} = a$,we substitute these into the equation:$2(s-2)v + 2v \cdot a = 0$Dividing by $2v$ (assuming $v \neq 0$):$(s-2) + a = 0$$a = -(s-2) = 2-s$At the point $(s, v) = (2-\sqrt{2}, \sqrt{2})$:$a = 2 - (2-\sqrt{2}) = \sqrt{2} \ ms^{-2}$

Class 11 Physics Motion in Straight Line Instantaneous Velocity and Speed and Velocity-time Graph

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The velocity-displacement $(v-s)$ graph shows the motion of a particle moving in a straight line. The velocity-displacement graph is a circle of radius $2 \ m$ and the center is at $(2, 0) \ m$. The value of acceleration for this particle at a point $(2-\sqrt{2}, \sqrt{2}) \ m$ will be $ms^{-2}$.

AP EAMCET 2018 All AP EAMCET

A
$\sqrt{2}$
B
$4$
C
$2$
D
$3$

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Motion in Straight Line

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Instantaneous Velocity and Speed and Velocity-time Graph

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$A$ particle covers a distance from $A$ to $B$ over a period of time; the distance versus time plot is shown below. Then which of the following is true for the motion of the particle?

EasyTS EAMCET 2019

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The velocity $(v)$-time $(t)$ graph for a particle moving along the $x$-axis is shown in the figure. The corresponding position $(x)$-time $(t)$ graph is best represented by:

Medium

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The figure shows a velocity-time graph of a particle moving along a straight line. The correct acceleration-time graph of the particle is shown as:

Medium

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The displacement-time graph of a moving particle is shown below. The instantaneous velocity of the particle is negative at the point:

EasyAIPMT 1994

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Given below are two statements:
Statement $I:$ Area under velocity-time graph gives the displacement of the body in a given time.
Statement $II:$ Area under acceleration-time graph is equal to the change in velocity in the given time.
In the light of given statements,choose the correct answer from the options given below.

MediumJEE MAIN 2023

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The velocity-displacement $(v-s)$ graph shows the motion of a particle moving in a straight line. The velocity-displacement graph is a circle of radius $2 \ m$ and the center is at $(2, 0) \ m$. The value of acceleration for this particle at a point $(2-\sqrt{2}, \sqrt{2}) \ m$ will be $ms^{-2}$.

Similar Questions

$A$ particle covers a distance from $A$ to $B$ over a period of time; the distance versus time plot is shown below. Then which of the following is true for the motion of the particle?

The velocity $(v)$-time $(t)$ graph for a particle moving along the $x$-axis is shown in the figure. The corresponding position $(x)$-time $(t)$ graph is best represented by:

The figure shows a velocity-time graph of a particle moving along a straight line. The correct acceleration-time graph of the particle is shown as:

The displacement-time graph of a moving particle is shown below. The instantaneous velocity of the particle is negative at the point:

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