(A) Let $s_A$ be the displacement of object $A$ and $s_B$ be the displacement of object $B$ at time $t$.For object $A$,initial velocity $u_A = 0$ and

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(A) Let $s_A$ be the displacement of object $A$ and $s_B$ be the displacement of object $B$ at time $t$.For object $A$,initial velocity $u_A = 0$ and acceleration is $a$. Using the equation of motion $s = ut + (1/2)at^2$,we get $s_A = (1/2)at^2$.For object $B$,velocity is constant $v$. Thus,$s_B = vt$.Since both objects meet at the same position at time $t$,their displacements must be equal: $s_A = s_B$.Therefore,$(1/2)at^2 = vt$.Dividing both sides by $t$ (assuming $t \neq 0$),we get $(1/2)at = v$.Solving for $t$,we find $t = 2v/a$.

Class 11 Physics Motion in Straight Line Uniformly Accelerated Motion

Medium

Object $A$ starts from rest with a constant acceleration $a$. Object $B$ starts from the same position and moves in the same direction as $A$ with a constant velocity $v$. If both meet after time $t$,then $t =$

A
$2v/a$
B
$v/a$
C
$v/(2a)$
D
$\sqrt{v/(2a)}$

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Class 11 Questions

Class 11

Physics Questions

Chapter

Motion in Straight Line

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Uniformly Accelerated Motion

$A$ particle returns to the starting point after $10\,s$. If the rate of change of velocity during the motion is constant,then its location after $7\,s$ will be same as that after $...........\,s$.

Medium

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$A$ car accelerates from rest at a constant rate $\alpha$ for some time,after which it decelerates at a constant rate $\beta$ and comes to rest. If the total time elapsed is $t$,then the maximum velocity acquired by the car is

DifficultIIT 1978

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$A$ particle starts from the origin at time $t=0$ and moves in the positive $x$-direction. Its velocity $v$ varies with time as $v=10t \text{ cm/s}$. The distance covered by the particle in $8 \text{ s}$ will be: (in $\text{ cm}$)

EasyTS EAMCET 2018

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$A$ car,starting from rest,accelerates at the rate $\alpha$ through a distance $d$,then continues at a constant speed for time $t$ and then decelerates at the rate of $\alpha/2$ to come to rest. If the total distance traveled is $15\, d$,then $d=$

Difficult

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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 ........$s$.

Difficult

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Object $A$ starts from rest with a constant acceleration $a$. Object $B$ starts from the same position and moves in the same direction as $A$ with a constant velocity $v$. If both meet after time $t$,then $t =$

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$A$ particle returns to the starting point after $10\,s$. If the rate of change of velocity during the motion is constant,then its location after $7\,s$ will be same as that after $...........\,s$.

$A$ car accelerates from rest at a constant rate $\alpha$ for some time,after which it decelerates at a constant rate $\beta$ and comes to rest. If the total time elapsed is $t$,then the maximum velocity acquired by the car is

$A$ particle starts from the origin at time $t=0$ and moves in the positive $x$-direction. Its velocity $v$ varies with time as $v=10t \text{ cm/s}$. The distance covered by the particle in $8 \text{ s}$ will be: (in $\text{ cm}$)

$A$ car,starting from rest,accelerates at the rate $\alpha$ through a distance $d$,then continues at a constant speed for time $t$ and then decelerates at the rate of $\alpha/2$ to come to rest. If the total distance traveled is $15\, d$,then $d=$

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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