JEE Main Motion in Straight Line PYQ

26. (JEE Main 2020 (Online) 8th January Evening Slot )

A particle moves such that its position vector $\vec{r} (t) = \cos ω t \hat{i} + \sin ω t \hat{j}$ where $ω$ is a constant and t is time. Then which of the following statements is true for the velocity $\vec{v} (t)$ and acceleration $\vec{a} (t)$ of the particle :

A. $\vec{v}$ and $\vec{a}$ both are perpendicular to $\vec{r}$

B. $\vec{v}$ and $\vec{a}$ both are parallel to $\vec{r}$

C. $\vec{v}$ is perpendicular to $\vec{r}$ and $\vec{a}$ is directed towards the origin

D. $\vec{v}$ is perpendicular to $\vec{r}$ and $\vec{a}$ is directed away from the origin

Correct Option is (C)

$\vec{r} (t) = \cos ω t \hat{i} + \sin ω t \hat{j}$

$\vec{v} = \frac{d \vec{r}}{d t}$ = $- ω \sin ω t \hat{i} + ω \cos ω t \hat{j}$

$\vec{a} = \frac{d \vec{v}}{d t}$ = $- ω^{2} \cos ω t \hat{i} - ω^{2} \sin ω t \hat{j}$

= $- ω^{2} (\cos ω t \hat{i} + \sin ω t \hat{j})$

= $- ω^{2} \vec{r}$

$∴$ $\vec{a}$ is antiparallel to $\vec{r}$ and it's direction towards the origin.

$\vec{v} . \vec{r} =$ $ω (- \sin ω t \cos ω t + \cos ω t \sin ω t)$ = 0

So $\vec{v} ⊥ \vec{r}$ .

27.(JEE Main 2020 (Online) 9th January Morning Slot )

The distance x covered by a particle in one dimensional motion varies with time t as
x² = at² + 2bt + c. If the acceleration of the particle depends on x as x^–n, where n is an integer, the value of n is __________

Correct answer is (3)

x² = at² + 2bt + c .....(1)

$\Rightarrow$ 2xv = 2at + 2b .....(2)

$\Rightarrow$ v = $\frac{a t + b}{x}$

differentiating (2) w.r.t. time

$\Rightarrow$ xa' + v² = a

Here a' is acceleration.

$\Rightarrow$ a'x = a - v² = a - ${[\frac{a t + b}{x}]}^{2}$

$\Rightarrow$ a'x = $\frac{a x^{2} - {(a t + b)}^{2}}{x^{2}}$

$\Rightarrow$ a' = $\frac{a (a t^{2} + 2 b t + c) - {(a t + b)}^{2}}{x^{2}}$

$\Rightarrow$ a' = $\frac{a c - b^{2}}{x^{3}}$

$∴$ a' $\propto$ $\frac{1}{x^{3}}$ $\propto$ x^-3

$∴$ n = 3

28. (JEE Main 2019 (Online) 12th April Evening Slot )

A particle is moving with speed v = b $\sqrt{x}$ along positive x-axis. Calculate the speed of the particle at time t = $τ$ (assume that the particle is at origin t = 0)

A. $\frac{b^{2} τ}{\sqrt{2}}$

B. $b^{2} τ$

C. $\frac{b^{2} τ}{2}$

D. $\frac{b^{2} τ}{4}$

Correct Option is (C)

v = b $\sqrt{x}$
$\Rightarrow$ $\frac{d x}{d t}$ = b $\sqrt{x}$
$\Rightarrow$ $\int_{0}^{x} \frac{d x}{\sqrt{x}} = \int_{0}^{t} b d t$
$\Rightarrow$ ${[\frac{x^{- \frac{1}{2} + 1}}{- \frac{1}{2} + 1}]}_{0}^{x}$ = $b {[t]}_{0}^{t}$
$\Rightarrow$ $x^{\frac{1}{2}} = \frac{b t}{2}$
$\Rightarrow$ $x = \frac{b^{2} t^{2}}{4}$

$∴$ v = $\frac{d x}{d t}$ = $\frac{b^{2}}{4}$ $\times$ 2t = $\frac{b^{2} t}{2}$
When t = $τ$ then speed v $= \frac{b^{2} τ}{2}$

29. (JEE Main 2019 (Online) 9th April Evening Slot )

The position of a particle as a function of time t, is given by
x(t) = at + bt² – ct³
where a, b and c are constants. When the particle attains zero acceleration, then its velocity will be :

A. $a + \frac{b^{2}}{c}$

B. $a + \frac{b^{2}}{4 c}$

C. $a + \frac{b^{2}}{3 c}$

D. $a + \frac{b^{2}}{2 c}$

Correct Option is (C)

x = at + bt² – ct³

$V = \frac{d x}{d t} = a + 2 b t - 3 c t^{2}$

$a = \frac{d v}{d t} = 2 b - 6 c t$

Put acceleration = 0

$\Rightarrow t = \frac{b}{3 c}$

Now V at t = $\frac{b}{3 c}$

$V = a + \frac{b^{2}}{3 c}$

30. (JEE Main 2018 (Online) 15th April Evening Slot )

A man in a car at location Q on a straight highway is moving with speed $υ$ . He decides to reach a point P in a field at a distance d from the highway (point M) as shown in the figure. Speed of the car in the field is half to that on the highway. What should be the distance RM, so that the time taken to reach P is minimum ?

JEE Main 2018 (Online) 15th April Evening Slot Physics - Motion Question 132 English

A. d

B. $\frac{d}{\sqrt{2}}$

C. $\frac{d}{2}$

D. $\frac{d}{\sqrt{3}}$

Correct Option is (D)

Let the distance QM = l and distance RM = x.

Time to reach from Q to R is $t_{1} = \frac{l - x}{v}$

Time to reach from R to P is $t_{2} = \frac{\sqrt{x^{2} + d^{2}}}{v / 2}$

Therefore, $t = t_{1} + t_{2} = \frac{l - x}{v} + \frac{\sqrt{x^{2} + d^{2}}}{v / 2}$

On differentiating, we get

$\frac{d t}{d x} = \frac{0 - 1}{v} + \frac{1}{v / 2} \frac{1}{2 \sqrt{x^{2} + d^{2}}} \times 2 x$

$\Rightarrow \frac{d t}{d x} = \frac{- 1}{v} + \frac{2 x}{v \sqrt{x^{2} + d^{2}}}$

JEE Main 2018 (Online) 15th April Evening Slot Physics - Motion Question 132 English Explanation

Put $\frac{d t}{d x} = 0$ , we get

$\frac{- 1}{v} + \frac{2 x}{v \sqrt{x^{2} + d^{2}}} = 0$

$\Rightarrow \frac{2 x}{v \sqrt{x^{2} + d^{2}}} = \frac{1}{v}$

$\Rightarrow 2 x = \sqrt{x^{2} + d^{2}} \Rightarrow 4 x^{2} = x^{2} + d^{2}$

$\Rightarrow 3 x^{2} = d^{2} \Rightarrow x = \frac{d}{\sqrt{3}}$

Therefore, the distance $R M = x = \frac{d}{\sqrt{3}}$ , the time taken to reach P is minimum.