### JEE Mains Previous Years Questions with Solutions

4.5
star star star star star
1

### 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 )
2

### 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.
3

### 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 ${t_1} = {t \over 6}$

$\Rightarrow S = {1 \over 2}f{\left( {{t \over 6}} \right)^2} = {{f\,{t^2}} \over {72}}$
4

### AIEEE 2004

A ball is thrown from a point with a speed ν0 at an angle of projection θ. From the same point and at the same instant person starts running with a constant speed ${{{v_0}} \over 2}$ to catch the ball. Will the person be able to catch the ball? If yes, what should be the angle of projection θ?
A
No
B
Yes, $30^\circ$
C
Yes, $60^\circ$
D
Yes, $45^\circ$

## Explanation

Yes, the person can catch the ball when horizontal velocity is equal to the horizontal component of ball's velocity, the motion of ball will be only in vertical direction with respect to person for that,

${{{v_0}} \over 2} = {v_0}\cos \theta \,\,\,\,$

or $\cos \theta = {1 \over 2}$

$\Rightarrow \cos \theta = \cos 60^\circ$

$\Rightarrow \theta = 60^\circ$