### JEE Mains Previous Years Questions with Solutions

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1

### AIEEE 2006

A particle located at x = 0 at time t = 0, starts moving along the positive x-direction with a velocity 'v' that varies as $v = \alpha \sqrt x$. The displacement of the particle varies with time as
A
t2
B
t
C
t1/2
D
t3

## Explanation

Given $v = \alpha \sqrt x$

$\therefore$ ${{dx} \over {dt}} = \alpha \sqrt x$

$\Rightarrow {{dx} \over {\sqrt x }} = \alpha \,dt$

On integrating we get,

$\int\limits_0^x {{{dx} \over {\sqrt x }} = \alpha \int\limits_0^t {dt} }$

$\Rightarrow \left[ {{{2\sqrt x } \over 1}} \right]_0^x = \alpha \left[ t \right]_0^t$

$\Rightarrow 2\sqrt x = \alpha t$

$\Rightarrow x = {{{\alpha ^2}} \over 4}{t^2}$

So $x \propto {t^2}$
2

### AIEEE 2005

The relation between time t and distance x is t = ax2 + bx where a and b are constants. The acceleration is
A
2bv3
B
-2abv2
C
2av2
D
-2av3

## Explanation

Given $t = a{x^2} + bx;$

Differentiate with respect to time $(t)$

${d \over {dt}}\left( t \right) = a{d \over {dt}}\left( {{x^2}} \right) + b{{dx} \over {dt}}$

$= a.2x{{dx} \over {dt}} + b.{{dx} \over {dt}}$

$1 = 2axv + bv = v\left( {2ax + b} \right)$

[as ${{dx} \over {dt}} = v$]

$\Rightarrow 2ax + b = {1 \over v}.$

Again differentiating,

$2a{{dx} \over {dt}} + 0 = - {1 \over {{v^2}}}{{dv} \over {dt}}$

$\Rightarrow {{dv} \over {dt}} = f = - 2a{v^3}$

(as ${{dv} \over {dt}} = f$ = Acceleration )
3

### AIEEE 2005

A particle is moving eastwards with a velocity of 5 m/s. In 10 seconds the velocity changes to 5 m/s northwards. The average acceleration in this time is
A
${1 \over 2}m{s^{ - 2}}$ towards north
B
${1 \over {\sqrt 2 }}m{s^{ - 2}}$ towards north-east
C
${1 \over {\sqrt 2 }}m{s^{ - 2}}$ towards north-west
D
zero

## Explanation

Average acceleration

$= {{change\,\,in\,\,velocioty} \over {time\,\,{\mathop{\rm int}} erval}}$

$= {{\Delta \overrightarrow v } \over t}$

${\overrightarrow v _1} = +5\widehat i$, towards east direction.

$\overrightarrow {{v_2}} = +5\widehat j$, towards north direction

$\Delta \overrightarrow v = \overrightarrow {{v_2}} - \overrightarrow {{v_1}}$ = $5\widehat i - 5\widehat j$

$\therefore$ $\,\,\,\,\overrightarrow a = {{5\widehat j - 5\widehat i} \over {10}} = {{\widehat j - \widehat i} \over 2}$

$\therefore$ $\,\,\,\, a = {{\sqrt {{1^2} + {{\left( { - 1} \right)}^2}} } \over 2} = {{\sqrt 2 } \over 2} = {1 \over {\sqrt 2 }}m{s^{ - 2}}$

$\tan \theta = {{{v_2}} \over {{v_1}}} = {5 \over 5} = 1$

$\therefore$ $\,\,\,\,\theta = {45^ \circ }$

Therefore the direction is North-west.
4

### AIEEE 2005

A car starting from rest accelerates at the rate f through a distance S, then continues at constant speed for time t and then decelerates at the rate ${f \over 2}$ to come to rest. If the total distance traversed is 15 S, then
A
$S = {1 \over 6}f{t^2}$
B
$S = ft$
C
$S = {1 \over 4}f{t^2}$
D
$S = {1 \over 72}f{t^2}$

## Explanation

Initially car starts from rest so u = 0.

Now distance from $A$ to $B$,

$\,\,\,\,\,\,\,\,\,\,$ $S = {1 \over 2}ft_1^2$

$\Rightarrow ft_1^2 = 2S$

Distance from $B$ to $C$ $= \left( {f{t_1}} \right)t$

In B to C velocity is constant and v = ${f{t_1}}$

Distance from $C$ to $D$
$\,\,\,\,\,\,\,\,\,\,$ $= {{{u^2}} \over {2a}} = {{{{\left( {f{t_1}} \right)}^2}} \over {2\left( {f/2} \right)}} = ft_1^2 = 2S$

$\Rightarrow S + f\,{t_1}t + 2S = 15S$

$\Rightarrow f\,{t_1}t = 12S$ ........(1)

But$\,\,\,\,\,\,\,\,\,$ ${1 \over 2}f\,t_1^2 = S$ .........(2)

On dividing the above two equations, we get ${t_1} = {t \over 6}$

$\Rightarrow S = {1 \over 2}f{\left( {{t \over 6}} \right)^2} = {{f\,{t^2}} \over {72}}$