JEE Main Motion in Straight Line PYQ

6. (JEE Main 2024 (Online) 1st February Evening Shift )

A particle initially at rest starts moving from reference point $x = 0$ along $x$ -axis, with velocity $v$ that varies as $v = 4 \sqrt{x} m / s$ . The acceleration of the particle is __________ ${ms}^{- 2}$ .

Correct answer is 8

To find the acceleration of the particle, we first need to differentiate the velocity function with respect to time. The velocity function given is

$v = 4 \sqrt{x}$

However, this function gives the velocity as a function of position $x$ , not as a function of time $t$ . Since acceleration is the rate of change of velocity with respect to time, we'll need to use the chain rule to differentiate $v$ with respect to $t$ .

The chain rule in this context can be stated as follows:

$a = \frac{d v}{d t} = \frac{d v}{d x} \cdot \frac{d x}{d t}$

Now, because $\frac{d x}{d t}$ is the velocity $v$ itself and $\frac{d v}{d x}$ is the derivative of the velocity with respect to $x$ , we first find $\frac{d v}{d x}$ :

$v = 4 \sqrt{x} = 4 x^{\frac{1}{2}}$

Differentiating with respect to $x$ , we get:

$\frac{d v}{d x} = 4 \cdot \frac{1}{2} x^{- \frac{1}{2}} = 2 x^{- \frac{1}{2}} = \frac{2}{\sqrt{x}}$

Now, because $v = 4 \sqrt{x}$ , we can rewrite $\sqrt{x}$ as $\frac{v}{4}$ . Using this to replace $\sqrt{x}$ in our expression for $\frac{d v}{d x}$ , we get:

$\frac{d v}{d x} = \frac{2}{\sqrt{x}} = \frac{2}{\frac{v}{4}} = \frac{8}{v}$

Now, using the chain rule:

$a = \frac{d v}{d t} = \frac{d v}{d x} \cdot \frac{d x}{d t} = \frac{8}{v} \cdot v$

Simplifying this, the velocity terms cancel out, leaving us with:

$a = 8 {ms}^{- 2}$

Thus, the acceleration of the particle is $8 {ms}^{- 2}$ .

7. (JEE Main 2024 (Online) 1st February Morning Shift )

A particle is moving in one dimension (along $x$ axis) under the action of a variable force. It's initial position was $16 m$ right of origin. The variation of its position $(x)$ with time $(t)$ is given as $x = - 3 t^{3} + 18 t^{2} + 16 t$ , where $x$ is in $m$ and $t$ is in $s$ .

The velocity of the particle when its acceleration becomes zero is _________ $m / s$ .

Correct answer is 52

Position (x): The particle's location on the x-axis at a given time.

Velocity (v): The rate of change of position with respect to time

( $v = \frac{d x}{d t}$ ).

Acceleration (a): The rate of change of velocity with respect to time

( $a = \frac{d v}{d t}$ ).

The particle's position is given by:

$x (t) = - 3 t^{3} + 18 t^{2} + 16 t$

We need to find the velocity when the acceleration is zero.

Find the Velocity (v) and Acceleration (a) Functions:

Velocity:

$v (t) = \frac{d x}{d t} = - 9 t^{2} + 36 t + 16$

Acceleration:

$a (t) = \frac{d v}{d t} = - 18 t + 36$

Find the Time (t) When Acceleration is Zero:

$a (t) = 0$

$- 18 t + 36 = 0$

$t = 2 seconds$

Calculate the Velocity at t = 2 seconds:

$v (2) = - 9 (2)^{2} + 36 (2) + 16$

$v (2) = 52 m/s$

Answer:

The velocity of the particle when its acceleration becomes zero is 52 m/s.

8. ( JEE Main 2023 (Online) 15th April Morning Shift)

The position of a particle related to time is given by $x = (5 t^{2} - 4 t + 5) m$ . The magnitude of velocity of the particle at $t = 2 s$ will be :

A. $14 {ms}^{- 1}$

B. $16 {ms}^{- 1}$

C. $10 {ms}^{- 1}$

D. $06 {ms}^{- 1}$

Correct Option is (B)

The position of a particle as a function of time is given by $x = (5 t^{2} - 4 t + 5) m$ .

To find the magnitude of the velocity of the particle at $t = 2 s$ , we first need to find the velocity of the particle as a function of time.

The velocity $v$ is the time derivative of the position $x$ :

$v = \frac{d x}{d t}$

Taking the derivative of $x$ with respect to $t$ , we get:

$v = \frac{d x}{d t} = 10 t - 4 m / s$

Now we can find the velocity of the particle at $t = 2 s$ by plugging in $t = 2$ :

$v (2 s) = 10 (2) - 4 m / s = 16 m / s$

Therefore, the magnitude of the velocity of the particle at $t = 2 s$ is:

$| v (2 s) | = 16 m / s$

9. (JEE Main 2023 (Online) 13th April Evening Shift)

The distance travelled by an object in time $t$ is given by $s = (2.5) t^{2}$ . The instantaneous speed of the object at $t = 5 s$ will be:

A. $5 {ms}^{- 1}$

B. $12.5 {ms}^{- 1}$

C. $62.5 {ms}^{- 1}$

D. $25 {ms}^{- 1}$

Correct Option is (D)

The distance traveled by an object in time $t$ is given by the equation $s = (2.5) t^{2}$ . To find the instantaneous speed at a specific time, we need to find the first derivative of the distance function with respect to time, which gives us the velocity function:

$v (t) = \frac{d s}{d t}$

Differentiating the given equation with respect to $t$ :

$v (t) = \frac{d}{d t} (2.5) t^{2} = 2 (2.5) t = 5 t$

Now, we can find the instantaneous speed at $t = 5 s$ by plugging the value into the velocity function:

$v (5) = 5 (5) = 25 m / s$

The instantaneous speed of the object at $t = 5 s$ is $25 m / s$ .

10. (JEE Main 2023 (Online) 10th April Evening Shift)

A person travels $x$ distance with velocity $v_{1}$ and then $x$ distance with velocity $v_{2}$ in the same direction. The average velocity of the person is $v$ , then the relation between $v, v_{1}$ and $v_{2}$ will be.

A. $V = V_{1} + V_{2}$

B. $V = \frac{v_{1} + V_{2}}{2}$

C. $\frac{1}{v} = \frac{1}{v_{1}} + \frac{1}{v_{2}}$

D. $\frac{2}{V} = \frac{1}{v_{1}} + \frac{1}{v_{2}}$

Correct Option is (D)

The average velocity is defined as the total displacement divided by the total time. Here, the person travels the same distance $x$ twice, once with velocity $v_{1}$ and once with velocity $v_{2}$ .

The time to travel distance $x$ with velocity $v_{1}$ is $t_{1} = \frac{x}{v_{1}}$ , and the time to travel distance $x$ with velocity $v_{2}$ is $t_{2} = \frac{x}{v_{2}}$ .

The total time is then

$t = t_{1} + t_{2} = \frac{x}{v_{1}} + \frac{x}{v_{2}}$ .

The total displacement is $2 x$ . So, the average velocity $v$ is given by

$v = \frac{total displacement}{total time} = \frac{2 x}{\frac{x}{v_{1}} + \frac{x}{v_{2}}} = \frac{2}{\frac{1}{v_{1}} + \frac{1}{v_{2}}}$

Multiplying both sides by $2$ , we get

$\frac{2}{v} = \frac{1}{v_{1}} + \frac{1}{v_{2}}$