(D) Given the relation: $t = \sqrt{x} + 3$.Rearranging for $x$: $\sqrt{x} = t - 3$.Squaring both sides: $x = (t - 3)^2$.Velocity $v$ is the derivative

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(D) Given the relation: $t = \sqrt{x} + 3$.Rearranging for $x$: $\sqrt{x} = t - 3$.Squaring both sides: $x = (t - 3)^2$.Velocity $v$ is the derivative of position with respect to time: $v = \frac{dx}{dt} = \frac{d}{dt}(t - 3)^2 = 2(t - 3)$.When the velocity is zero: $2(t - 3) = 0$,which gives $t = 3 \ s$.Initial position at $t = 0 \ s$: $x_i = (0 - 3)^2 = 9 \ m$.Final position at $t = 3 \ s$: $x_f = (3 - 3)^2 = 0 \ m$.Displacement is defined as the change in position: $\Delta x = x_f - x_i = 0 - 9 = -9 \ m$.

Class 11 Physics Motion in Straight Line Uniformly Accelerated Motion

Medium

The relation between the position $x$ and time $t$ of a particle moving in one dimension is given by $t = \sqrt{x} + 3$,where $x$ is in meters and $t$ is in seconds. When the velocity of the particle becomes zero,the displacement of the particle is ........ $m$.

A
$0$
B
$3$
C
$-3$
D
$-9$

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Motion in Straight Line

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Uniformly Accelerated Motion

$A$ body starts from rest with uniform acceleration $a$. Its velocity after $n \ s$ is $v$. The displacement of the body in the last $3 \ s$ is: (Assume total time of journey is $n \ s$)

Medium

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$A$ car is moving along a straight road with uniform acceleration. It passes through two points $P$ and $Q$ separated by a distance $s$ with velocities $30\; km/h$ and $40\; km/h$ respectively. Find the velocity of the car midway between $P$ and $Q$.

MediumAIPMT 1988

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Two cars $A$ and $B$ are initially at rest at the same point. If car $A$ starts with a uniform velocity of $40\, m/s$ and car $B$ starts in the same direction with a constant acceleration of $4\, m/s^2$,then after how much time will car $B$ catch car $A$?

Medium

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$A$ man of height $h$ is walking away from a street lamp with a constant speed $v$. The height of the street lamp is $3h$. The rate at which the length of the man's shadow is increasing when he is at a distance $10h$ from the base of the street lamp is:

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$A$ rifle bullet loses $1/20^{th}$ of its velocity in passing through a wooden plank. The least number of planks required to stop the bullet is :-

Difficult

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The relation between the position $x$ and time $t$ of a particle moving in one dimension is given by $t = \sqrt{x} + 3$,where $x$ is in meters and $t$ is in seconds. When the velocity of the particle becomes zero,the displacement of the particle is ........ $m$.

Similar Questions

$A$ body starts from rest with uniform acceleration $a$. Its velocity after $n \ s$ is $v$. The displacement of the body in the last $3 \ s$ is: (Assume total time of journey is $n \ s$)

$A$ car is moving along a straight road with uniform acceleration. It passes through two points $P$ and $Q$ separated by a distance $s$ with velocities $30\; km/h$ and $40\; km/h$ respectively. Find the velocity of the car midway between $P$ and $Q$.

Two cars $A$ and $B$ are initially at rest at the same point. If car $A$ starts with a uniform velocity of $40\, m/s$ and car $B$ starts in the same direction with a constant acceleration of $4\, m/s^2$,then after how much time will car $B$ catch car $A$?

$A$ man of height $h$ is walking away from a street lamp with a constant speed $v$. The height of the street lamp is $3h$. The rate at which the length of the man's shadow is increasing when he is at a distance $10h$ from the base of the street lamp is:

$A$ rifle bullet loses $1/20^{th}$ of its velocity in passing through a wooden plank. The least number of planks required to stop the bullet is :-

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