1

### JEE Main 2019 (Online) 10th January Evening Slot

A particle starts from the origin at time t = 0 and moves along the positive x-axis. The graph of velocity with respect to time is shown in figure. What is the position of the particle at time t = 5s ? A
3 m
B
9 m
C
10 m
D
6 m

## Explanation

S = Area under graph

${1 \over 2}$ $\times$ 2$\times$2 + 2 $\times$ 2 + 3 $\times$ 1 = 9 m
2

### JEE Main 2019 (Online) 11th January Morning Slot

A body is projected at t = 0 with a velocity 10 ms–1 at an angle of 60o with the horizontal. The radius of curvature of its trajectory at t = 1s is R. neglecting air resistance and taking acceleration due to gravity g = 10 ms–2, the value of R is :
A
2.8 m
B
5.1 m
C
2.5 m
D
10.3 m

## Explanation vx = 10cos60o = 5 m/s vy = 10cos30o = $5\sqrt 3$ m/s

velocity after t = 1 sec.

vx = 5 m/s

vy = $\left| {\left( {5\sqrt 3 - 10} \right)} \right|$ m/s = 10 $-$ 5$\sqrt 3$

an = ${{{v^2}} \over R} \Rightarrow \,R\,$ = ${{v_x^2 + v_y^2} \over {{a_n}}}$ = ${{25 + 100 + 75 - 100\sqrt 3 } \over {10\cos \theta }}$

tan$\theta$ = ${{10 - 5\sqrt 3 } \over 5}$ = 2 $-$ ${\sqrt 3 }$ $\Rightarrow$  $\theta$ = 15o

R = ${{100\left( {2 - \sqrt 3 } \right)} \over {10\cos 15}} = 2.8m$
3

### JEE Main 2019 (Online) 11th January Evening Slot

A particle moves from the point $\left( {2.0\widehat i + 4.0\widehat j} \right)$ m, at t = 0, with an initial velocity $\left( {5.0\widehat i + 4.0\widehat j} \right)$ ms$-$1. It is acted upon by a constant force which produces a constant acceleration $\left( {4.0\widehat i + 4.0\widehat j} \right)$ ms$-$2. What is the distance of the particle from the origin at time 2 s?
A
15 m
B
$20\sqrt 2$ m
C
$10\sqrt 2$ m
D
5 m

## Explanation

$\overrightarrow S = \left( {5\widehat i + 4} \right)2 + {1 \over 2}\left( {4\widehat i + 4\widehat j} \right)4$

$= 10\widehat i + 8\widehat j + 8\widehat i + 8\widehat j$

$\overrightarrow {{r_f}} - \overrightarrow {{r_i}} = 18\widehat i + 16\widehat j$

$\overrightarrow {{r_f}} = 20\widehat i + 20\widehat j$

$\left| {\overrightarrow {{r_f}} } \right| = 20\sqrt 2$
4

### JEE Main 2019 (Online) 12th January Morning Slot

A person standing on an open ground hears the sound of a jet aeroplane, coming from north at an angle 60o with ground level. But he finds the aeroplane right vertically above his position. If v is the speed of sound, speed of the plane is :
A
${{\sqrt 3 } \over 2}$v
B
${{2v} \over {\sqrt 3 }}$
C
v
D
${v \over 2}$

## Explanation AB = VP $\times$ t

BC = Vt

cos60o = ${{AB} \over {BC}}$

${1 \over 2} = {{{V_P} \times t} \over {Vt}}$

VP = ${V \over 2}$