JEE Main Motion in Straight Line PYQ

16. (JEE Main 2021 (Online) 25th February Morning Shift )

An engine of a train, moving with uniform acceleration, passes the signal-post with velocity u and the last compartment with velocity v. The velocity with which middle point of the train passes the signal post is :

A. $\frac{u + v}{2}$

B. $\sqrt{\frac{v^{2} - u^{2}}{2}}$

C. $\frac{v - u}{2}$

D. $\sqrt{\frac{v^{2} + u^{2}}{2}}$

Correct Option is (D)

JEE Main 2021 (Online) 25th February Morning Shift Physics - Motion Question 98 English Explanation
Let initial speed of train u. When midpoint of the train reach the signal post it's velocity becomes v₀.

$∴$ $v_{0}^{2} = u^{2} + 2 a s$ .......(1)

When train passes the signal post completely it's velocity becomes v.

$∴$ $v^{2} = v_{0}^{2} + 2 a s$ ......(2)

Subtracting (2) from (1) we get,

$v_{0}^{2} - v^{2} = u^{2} - v_{0}^{2}$

$\Rightarrow v_{0}^{2} + v_{0}^{2} = u^{2} + v^{2}$

$\Rightarrow v_{0} = \sqrt{\frac{u^{2} + v^{2}}{2}}$

17. (JEE Main 2020 (Online) 5th September Morning Slot )

A balloon is moving up in air vertically above a point A on the ground. When it is at a height h₁, a girl standing at a distanced (point B) from A (see figure) sees it at an angle 45^o with respect to the vertical. When the balloon climbs up a further height h₂, it is seen at an angle 60^o with respect to the vertical if the girl moves further by a distance 2.464 d(point C). Then the height h₂ is (given tan 30^o = 0.5774) JEE Main 2020 (Online) 5th September Morning Slot Physics - Motion Question 102 English

A. 0.464d

B. d

C. 0.732d

D. 1.464d

Correct Option is (B)

JEE Main 2020 (Online) 5th September Morning Slot Physics - Motion Question 102 English Explanation

From $Δ A B D$

$\tan 45 = \frac{h_{1}}{d}$

$\Rightarrow 1 = \frac{h_{1}}{d}$

$\Rightarrow h_{1} = d$

From $Δ A C E$

$\tan 30 = \frac{h_{1} + h_{2}}{d + 2.464 d}$

$\Rightarrow$ $0.5774 = \frac{d + h_{2}}{3.464 d}$

$\Rightarrow$ $d + h_{2} = 0.5774 \times 3.464 \times d$

$\Rightarrow$ $h_{2} = 2.0001136 d - d$

$\Rightarrow$ $h_{2} = 2.000 d - d = d$

18. (JEE Main 2020 (Online) 8th January Morning Slot )

A particle is moving along the x-axis with its coordinate with the time 't' given be
x(t) = 10 + 8t – 3t². Another particle is moving the y-axis with its coordinate as a function of time given by y(t) = 5 – 8t³.
At t = 1s, the speed of the second particle as measured in the frame of the first particle is given as $\sqrt{v}$ . Then v (in m/s) is ______.

Correct answer is (580)

For particle ‘A’

x = 10 + 8t – 3t²

$\Rightarrow$ v_A = 8 – 6t

at t = 1 s

$\vec{v_{A}}$ = 2 $\hat{i}$

For particle ‘B’

y = 5 – 8t³

$\Rightarrow$ v_B = – 24t²

at t = 1 s

$\vec{v_{B}}$ = -24 $\hat{j}$

Velocity of B with respect to A

$\vec{v_{B / A}}$ = $\vec{v_{B}}$ - $\vec{v_{A}}$

= -24 $\hat{j}$ - 2 $\hat{i}$

$∴$ $| \vec{v_{B / A}} |$ = $\sqrt{{(24)}^{2} + {(2)}^{2}}$

= $\sqrt{580}$

$∴$ v = 580

19. (JEE Main 2019 (Online) 9th January Evening Slot )

In a car race on straight road, car A takes a time t less than car B at the finish and passes finishing point with a speed ' $υ$ ' more than that of car B. Both the cars start from rest and travel with constant acceleration a₁ and a₂ respectively. Then ' $υ$ ' is equal to :

A. $\frac{2 a_{1} a_{2}}{a_{1} + a_{2}} t$

B. $\sqrt{2 a_{1} a_{2}} t$

C. $\sqrt{a_{1} a_{2}} t$

D. $\frac{a_{1} + a_{2}}{2} t$

Correct Option is (C)

For both car initial speed ( $μ$ ) = 0

Let the acceleration of car A and car B is $a$ ₁ and $a$ ₂ respectively.

Also let the time taken to reach the finishing point for car A is t₁ and for car B is t₂.

Let at finishing point speed of car A is $v$ ₁ and speed of car B is $v$ ₂

According to the question,

t₂ $-$ t₁ = t

and    $v$ ₁ $-$ $v$ ₂ = $v$

$\Rightarrow$    $a$ ₁t₁ $-$ $a$ ₂t₂ = $v$

$\Rightarrow$    $a$ ₁t₁ $-$ $a$ ₂(t + t₁) = $v$ . . . . . . .(1)

As, Total distance covered by both car is equal.

So,   x_A = x_B

$\Rightarrow$    $\frac{1}{2} a_{1} t_{1}^{2} = \frac{1}{2} a_{2} t_{2}^{2}$

$\Rightarrow$    $a$ ₁t $_{1}^{2}$ = $a$ ₂ (t + t₁)²

$\Rightarrow$    $\sqrt{a_{1}} . t_{1}$ = $\sqrt{a_{2}}$ . (t + t₁)

$\Rightarrow$    $\sqrt{a_{1}} . t_{1} - \sqrt{a_{2}} . t_{1} = \sqrt{a_{2}} . t$

$\Rightarrow$   t₁ = $\frac{\sqrt{a_{2}} . t}{\sqrt{a_{1}} - \sqrt{a_{2}}} . . . . . (2)$

Now put the value of t₁ in equation (2),

( $a$ ₁ $-$ $a$ ₂) t₁ $-$ $a$ ₂t = $v$

$\Rightarrow$   (a₁ $-$ a₂) . $\frac{\sqrt{a_{2}} . t}{\sqrt{a_{1}} - \sqrt{a_{2}}} - a_{2} t = v$

$\Rightarrow$    $(\sqrt{a_{1}} + \sqrt{a_{2}}) \sqrt{a_{2}} . t - a_{2} t = v$

$\Rightarrow$    $\sqrt{a_{1} a_{2}} . t + a_{2} . t - a_{2} t = v$

$\Rightarrow$    $v$ = $\sqrt{a_{1} a_{2}} . t$

20. (JEE Main 2018 (Online) 15th April Morning Slot )

An automobile, travelling at $40$ km/h, can be stopped at a distance of $40$ m by applying brakes. If the same automobile is travelling at $80$ km/h, the minimum stopping distance, in metres, is (assume no skidding) :

A. $45$ m

B. $100$ m

C. $150$ m

D. $160$ m

Correct Option is (D)

In first case

u₁ = 40 km / h
$υ$ ₁ = 0
s₁ = 40 m

using, $υ$ ² $-$ $μ$ ² = 2 as

0² $-$ 40² = 2a $\times$ 40 . . . . . .(1)

In 2nd case

0² $-$ 80² = 2 as . . . . . .(2)

dividing (2) by (1) we get,

$\frac{s}{40}$ = $\frac{80^{2}}{40^{2}}$

$\Rightarrow$ S = $\frac{80 \times 80}{40}$

$\Rightarrow$ S = 160 m