(A) Given the velocity equation: $v = \sqrt{2x + 9}$.Squaring both sides,we get: $v^2 = 2x + 9$.Comparing this with the kinematic equation $v^2 = u^2

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$1\, \text{unit}, 3\, \text{unit}$

(A) Given the velocity equation: $v = \sqrt{2x + 9}$.Squaring both sides,we get: $v^2 = 2x + 9$.Comparing this with the kinematic equation $v^2 = u^2 + 2ax$,where $u$ is the initial velocity and $a$ is the acceleration:At the origin,$x = 0$. Substituting $x = 0$ into the velocity equation: $v = \sqrt{2(0) + 9} = \sqrt{9} = 3\, \text{unit}$. Thus,the initial velocity $u = 3\, \text{unit}$.Comparing $v^2 = 2x + 9$ with $v^2 = u^2 + 2ax$,we identify $2a = 2$,which gives $a = 1\, \text{unit}$.Therefore,the initial acceleration is $1\, \text{unit}$ and the initial velocity is $3\, \text{unit}$.

Class 11 Physics Motion in a Straight Line Uniformly Accelerated Motion

Medium

The position $x$ of a particle at any instant is related to its velocity $v$ by the equation $v = \sqrt{2x + 9}$. The particle starts from the origin. What are the initial acceleration and initial velocity of the particle?

A
$1\, \text{unit}, 3\, \text{unit}$
B
$2\, \text{unit}, 3\, \text{unit}$
C
$4\, \text{unit}, 81\, \text{unit}$
D
$9\, \text{unit}, 3\, \text{unit}$

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Motion in a Straight Line

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Uniformly Accelerated Motion

$A$ body starts from rest with uniform acceleration and its velocity at a time of $n$ seconds is $v$. The total displacement of the body in the $n^{\text{th}}$ and $(n-1)^{\text{th}}$ seconds of its motion is:

MediumAP EAMCET 2025

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$A$ motorist starting a car from rest accelerates uniformly to a speed of $v \ m/s$ in $9 \ s$. He maintains this speed for another $50 \ s$ and then applies the brakes and decelerates uniformly to rest. His deceleration is numerically equal to three times his previous acceleration. Then the time during which the deceleration takes place is .......... $s$ :-

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$A$ particle moves along a straight line $OX$. At a time $t$ (in seconds),the distance $x$ (in metres) of the particle from $O$ is given by $x = 40 + 12t - t^3$. How long would the particle travel before coming to rest?

MediumAIPMT 2006

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$A$ car starts at time $t=0$ from an initial speed of $10 \,m/s$ and accelerates uniformly with $2 \,m/s^2$ on a straight road for time $0 \leq t \leq 10 \,s$. Let $S_1$ and $S_2$ be the distance covered by the car in time $3 \leq t \leq 4 \,s$ and $4 \leq t \leq 5 \,s$ respectively. The ratio $\frac{S_2}{S_1}$ is

EasyTS EAMCET 2022

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The velocity of a bullet is reduced from $200\,m/s$ to $100\,m/s$ while travelling through a wooden block of thickness $10\,cm$. Assuming the retardation to be uniform,the retardation will be:

EasyAIIMS 2001

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The position $x$ of a particle at any instant is related to its velocity $v$ by the equation $v = \sqrt{2x + 9}$. The particle starts from the origin. What are the initial acceleration and initial velocity of the particle?

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$A$ body starts from rest with uniform acceleration and its velocity at a time of $n$ seconds is $v$. The total displacement of the body in the $n^{\text{th}}$ and $(n-1)^{\text{th}}$ seconds of its motion is:

$A$ particle moves along a straight line $OX$. At a time $t$ (in seconds),the distance $x$ (in metres) of the particle from $O$ is given by $x = 40 + 12t - t^3$. How long would the particle travel before coming to rest?

The velocity of a bullet is reduced from $200\,m/s$ to $100\,m/s$ while travelling through a wooden block of thickness $10\,cm$. Assuming the retardation to be uniform,the retardation will be:

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