JEE Main Motion in Straight Line PYQ

16.(JEE Main 2023 (Online) 29th January Morning Shift )

A tennis ball is dropped on to the floor from a height of 9.8 m. It rebounds to a height 5.0 m. Ball comes in contact with the floor for 0.2s. The average acceleration during contact is ___________ ms $^{- 2}$ .

(Given g = 10 ms $^{- 2}$ )

Correct answer is (120)

The speed of ball just before collision with ground is

$u = \sqrt{2 \times g H} = \sqrt{2 \times 10 \times 9.8} = \underset{(Downwards)}{14 m / \sec}$

The speed of ball just after collision is

$v = \sqrt{2 g h} = \sqrt{2 \times 10 \times 5} = \underset{(Upwards)}{10 m / \sec}$

So, $\vec{a} = \frac{Δ \vec{v}}{Δ t}$

$= \frac{10 + 14}{0.2} = 120 m / s^{2}$

17. (JEE Main 2022 (Online) 29th July Morning Shift )

If $t = \sqrt{x} + 4$ , then ${(\frac{d x}{d t})}_{t = 4}$ is :

A. 4

B. Zero

C. 8

D. 16

Correct Option is (B)

Given,

$t = \sqrt{x} + 4$ , Squaring on both

$\begin{aligned} x = (t - 4)^{2} = t^{2} - 8 t + 16 \\ \frac{d x}{d t} = 2 t - 8 \\ at t = 4 \\ \frac{d x}{d t} = 8 - 8 = 0 \end{aligned}$

18. (JEE Main 2022 (Online) 28th June Evening Shift )

Velocity (v) and acceleration (a) in two systems of units 1 and 2 are related as $v_{2} = \frac{n}{m^{2}} v_{1}$ and $a_{2} = \frac{a_{1}}{m n}$ respectively. Here m and n are constants. The relations for distance and time in two systems respectively are :

A. $\frac{n^{3}}{m^{3}} L_{1} = L_{2}$ and $\frac{n^{2}}{m} T_{1} = T_{2}$

B. $L_{1} = \frac{n^{4}}{m^{2}} L_{2}$ and $T_{1} = \frac{n^{2}}{m} T_{2}$

C. $L_{1} = \frac{n^{2}}{m} L_{2}$ and $T_{1} = \frac{n^{4}}{m^{2}} T_{2}$

D. $\frac{n^{2}}{m} L_{1} = L_{2}$ and $\frac{n^{4}}{m^{2}} T_{1} = T_{2}$

Correct Option is (A)

$[L] = \frac{[v^{2}]}{[a]}$

so $\frac{{[v_{2}]}^{2}}{[a_{2}]} = \frac{{[\frac{n}{m^{2}} v_{1}]}^{2}}{[\frac{a_{1}}{m n}]}$

$\frac{{[v_{2}]}^{2}}{[a_{2}]} = \frac{n^{3}}{m^{3}} \frac{{[v_{1}]}^{2}}{[a_{1}]}$

or $[L_{2}] = \frac{n^{3}}{m^{3}} [L_{1}]$

Similarly,

$[T] = \frac{[v]}{[a]}$

So, $[T_{2}] = \frac{n^{2}}{m} [T_{1}]$

19. (JEE Main 2022 (Online) 25th June Evening Shift )

Two buses P and Q start from a point at the same time and move in a straight line and their positions are represented by $X_{P} (t) = α t + β t^{2}$ and $X_{Q} (t) = f t - t^{2}$ . At what time, both the buses have same velocity?

A. $\frac{α - f}{1 + β}$

B. $\frac{α + f}{2 (β - 1)}$

C. $\frac{α + f}{2 (1 + β)}$

D. $\frac{f - α}{2 (1 + β)}$

Correct Option is (D)

$X_{P} = α t + β t^{2}$

$X_{Q} = f t - t^{2}$

$∴$ $V_{P} = α + 2 β t$

$V_{Q} = f - 2 t$

$∵$ $V_{P} = V_{Q}$

$\Rightarrow α + 2 β t = f - 2 t$

$\Rightarrow t = \frac{f - α}{2 (1 + β)}$

20.(JEE Main 2022 (Online) 28th June Evening Shift )

A car covers AB distance with first one-third at velocity v₁ ms^{$-$ 1}, second one-third at v₂ ms^{$-$ 1} and last one-third at v₃ ms^{$-$ 1}. If v₃ = 3v₁, v₂ = 2v₁ and v₁ = 11 ms^{$-$ 1} then the average velocity of the car is _____________ ms^{$-$ 1}.

JEE Main 2022 (Online) 28th June Evening Shift Physics - Motion Question 64 English

Correct answer is (18)

$v_{m e a n} = \frac{3 v_{1} v_{2} v_{3}}{v_{1} v_{2} + v_{2} v_{3} + v_{3} v_{1}}$

$= \frac{3 \times 11 \times 22 \times 33}{11 \times 22 + 22 \times 33 + 33 \times 11}$

$= 18$ m/sec