(D) Using the third equation of motion: $v^2 = u^2 + 2as$.Since the final velocity $v = 0$,we have $0 = u^2 - 2as$,which implies $s = \frac{u^2}{2a}$.

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(D) Using the third equation of motion: $v^2 = u^2 + 2as$.Since the final velocity $v = 0$,we have $0 = u^2 - 2as$,which implies $s = \frac{u^2}{2a}$.Assuming the deceleration $a$ is constant,the stopping distance $s$ is directly proportional to the square of the initial velocity: $s \propto u^2$.Therefore,$\frac{s_2}{s_1} = \left( \frac{u_2}{u_1} \right)^2$.Given $u_1 = 30 \, km/hr$,$s_1 = 8 \, m$,and $u_2 = 60 \, km/hr$:$\frac{s_2}{8} = \left( \frac{60}{30} \right)^2 = (2)^2 = 4$.$s_2 = 4 \times 8 = 32 \, m$.

Class 11 Physics Motion in Straight Line Uniformly Accelerated Motion

Medium

$A$ car moving at a speed of $30 \, km/hr$ comes to a stop after traveling $8 \, m$ when the brakes are applied. If the same car is moving at a speed of $60 \, km/hr$,what distance (in $m$) will it travel before coming to a stop after the brakes are applied?

A
$8$
B
$16$
C
$24$
D
$32$

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Motion in Straight Line

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Uniformly Accelerated Motion

The relation between time $t$ and distance $x$ of a particle is $t = ax^2 + bx$, where $a$ and $b$ are constants. If $v$ is the velocity of the particle, then its acceleration is

EasyTS EAMCET 2023

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The displacement $x$ (in $m$) of a particle of mass $m$ (in $kg$) moving in one dimension under the action of a force,is related to time $t$ (in $s$) by $t = \sqrt{x} + 3$. The displacement of the particle when its velocity is zero,will be

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Two cars leave one after the other and travel with an acceleration of $0.4\, m/s^2$. Two minutes after the departure of the first,the distance between the cars becomes $1.9\, km$. The time interval between the departures of the cars is ........$s$.

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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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$A$ body moving with uniform acceleration travels a distance of $25 \ m$ in the fourth second and $37 \ m$ in the sixth second. The distance covered by the body in the next two seconds is: (in $m$)

MediumTS EAMCET 2024

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$A$ car moving at a speed of $30 \, km/hr$ comes to a stop after traveling $8 \, m$ when the brakes are applied. If the same car is moving at a speed of $60 \, km/hr$,what distance (in $m$) will it travel before coming to a stop after the brakes are applied?

Similar Questions

The relation between time $t$ and distance $x$ of a particle is $t = ax^2 + bx$, where $a$ and $b$ are constants. If $v$ is the velocity of the particle, then its acceleration is

The displacement $x$ (in $m$) of a particle of mass $m$ (in $kg$) moving in one dimension under the action of a force,is related to time $t$ (in $s$) by $t = \sqrt{x} + 3$. The displacement of the particle when its velocity is zero,will be

Two cars leave one after the other and travel with an acceleration of $0.4\, m/s^2$. Two minutes after the departure of the first,the distance between the cars becomes $1.9\, km$. The time interval between the departures of the cars is ........$s$.

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 $A$?

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